The x-axis point 2.202 m to the right of the origin is where the electric field caused by the two charges is zero.
A positive point charge is what?A positive line charge with a uniform density is separated by r0 from a positive point charge that has been released from rest. The speed v of the point charge is proportional to A as a function of the instantaneous separation r from the line charge (rr0)
The electric field caused by Q1 at a position along the x-axis is given by:
E1 = kQ1/x^2
Similar to that, the electric field caused by Q2 at a point on the x-axis is given by:
E2 = kQ2/(x - 2)^2
As a result, we must determine the value of x where:
E1 + E2 = 0
The result of substituting the values of E1 and E2 is:
kQ1/x^2 + kQ2/(x - 2)^2 = 0
Simplifying the equation, we get:
Q1/(4πε0x^2) + Q2/(4πε0(x - 2)^2) = 0
where ε0 is the permittivity of free space.
Substituting the given values of Q1 and Q2, we get:
(2.8×10−5)/(4πε0x^2) - (5.4×10−6)/(4πε0(x - 2)^2) = 0
Multiplying both sides by (4πε0x^2(x - 2)^2), we get:
(2.8×10−5)(x - 2)^2 - (5.4×10−6)x^2 = 0
Expanding the equation and simplifying, we get a quadratic equation in x:
2.8×10−5x^2 - 1.12×10−4x + 6.48×10−6 = 0
Solving this quadratic equation using the quadratic formula, we get:
x = 0.064 m or x = 2.202 m
The null point must be to the right of the positive point charge because the positive point charge is fixed at the origin. Hence, the null point is at x = 2.202 m.
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Four students each construct a claim about the gravitational attraction on a probe passing between Mercury and Venus when the planets are nearest in their orbits.
Student W claims that the probe is pulled harder by the planet with the greater mass when halfway between the planets.
Student X claims that the probe is pulled harder by the planet nearest the Sun when halfway between the planets.
Student Y claims that the probe is pulled harder by the planet with the greater mass anywhere between the planets.
Student Z claims that the probe is pulled harder by the planet nearest the Sun anywhere between the planets.
Which student made a claim that is best supported by evidence from the table and diagram?
Answer:
Without the table and diagram, it's difficult to provide a specific answer. However, in general, the gravitational attraction between two objects depends on their masses and the distance between them. The force of attraction is stronger when the masses are greater and when the objects are closer together.
Based on this understanding, we can evaluate the claims made by each student:
Student W claims that the probe is pulled harder by the planet with the greater mass when halfway between the planets. This claim is partially supported by the evidence, since the gravitational force is stronger when the masses are greater. However, it doesn't take into account the distance between the planets and the probe.
Student X claims that the probe is pulled harder by the planet nearest the Sun when halfway between the planets. This claim is not supported by the evidence, since the distance between the probe and each planet is not specified.
Student Y claims that the probe is pulled harder by the planet with the greater mass anywhere between the planets. This claim is partially supported by the evidence, since the gravitational force is stronger when the masses are greater. However, it doesn't take into account the distance between the planets and the probe.
Student Z claims that the probe is pulled harder by the planet nearest the Sun anywhere between the planets. This claim is not supported by the evidence, since the distance between the probe and each planet is not specified.
Overall, Student W and Student Y made claims that are partially supported by the evidence, but neither claim takes into account the distance between the planets and the probe. Therefore, it's difficult to determine which claim is best supported by the evidence without more information.
Explanation:
Please see Attached.
The values for the work done to the nearest 10 is as follows:
(a) WA = 13,200 J.(b) Wg = 0 J.(c) WN = 0 J.(d) WF = 3,550 J.(e) WNet = 16,750 J.(f) WNet = 16,750 J.What is the work done?(a) The work done by the applied force can be calculated using the formula:
WA = FAdcosθ,
where FA is the applied force, d is the distance moved, θ is the angle between the force and the displacement, and cosθ is the cosine of that angle.
In this case, FA = 216 N, d = 71.0 m, and θ = 30.0° below the horizontal. Therefore,
WA = (216 N)(71.0 m)cos(-30.0°)
WA = 13,200 J.
(b) The force of gravity on the box is given by Fg = mg
The work done by the force of gravity is given by Wg = Fgd,
Fg = (51.0 kg)(9.81 m/s^2)
Fg = 500.31 N, and the box does not move vertically, so d = 0.
Therefore, the work done by the force of gravity is zero.
Wg = 0 J.
(c) The normal force on the box is equal in magnitude and opposite in direction to the force of gravity, so FN = Fg. Since the box does not move vertically, the normal force does not do any work.
Therefore,
WN = 0 J.
(d) The force of friction on the box can be calculated using the formula:
FF = μFN,
where μ is the coefficient of kinetic friction between the floor and the box, and FN is the normal force on the box.
In this case, μ = 0.100 and FN = Fg = 500.31 N. Therefore,
FF = (0.100)(500.31 N) = 50.03 N
The work done by the force of friction can be calculated using the formula:
WF = FFd
In this case, the box moves a distance of 71.0 m.
Therefore,
WF = (50.03 N)(71.0 m)
WF = 3,550 J.
(e) The net work on the box is the sum of the works done by each individual force.
Therefore,
WNet = WA + Wg + WN + WF = 13,200 J + 0 J + 0 J + 3,550 J
WNet = 16,750 J.
(f) Alternatively, we can find the net work by first finding the net force on the box.
The horizontal component of the applied force is FAcosθ, where θ = 30.0° below the horizontal.
Therefore,
FNet,x = FAcosθ - FF = (216 N)cos(30.0°) - (0.100)(500.31 N)
FNet,x = 91.10 N
The vertical component of the applied force is FAsinθ, where θ = 30.0° below the horizontal. Therefore,
FNet,y = Fg - FAsinθ = (51.0 kg)(9.81 m/s^2) - (216 N)sin(30.0°)
FNet,y = 247.16 N
The net force on the box can be found using the Pythagorean theorem:
FNet = √(FNet,x² + FNet,y²)
FNet = √((91.10 N)² + (247.16 N)²)
FNet = 264.36 N
The direction of the net force is given by the angle θ between FNet,x and FNet:
θ = atan(FNet,y / FNet,x)
θ = atan(247.16 N / 91.10 N
θ = 70.9°
The net work done on the box is given by the formula:
WNet = FNetdcosθ, where d is the distance moved by the box.
In this case, d = 71.0 m.
Therefore,
WNet = (264.36 N)(71.0 m)cos(70.9°) = 16,750 J
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A 4.9-m wide swimming pool is filled to the top. The bottom of the pool becomes completely shaded in the afternoon when the sun is 25 degrees above the horizon. How deep is the pool?
Answer:
Explanation:
We can solve this problem using trigonometry. Let's draw a diagram:
|\
| \
| \ <- Sun rays
| \
| \
| \
| \
---------
Pool
The angle between the sun rays and the horizontal line is 90 - 25 = 65 degrees. Let's call the depth of the pool "d". We want to find the value of "d" that makes the bottom of the pool completely shaded.
We can see that the triangle formed by the sun rays, the top edge of the pool, and the bottom edge of the shaded area is a right triangle. The angle between the sun rays and the top edge of the pool is also 65 degrees, because the top edge is parallel to the ground.
Using trigonometry, we can write:
tan(65 degrees) = d / 4.9 m
Solving for "d", we get:
d = 4.9 m * tan(65 degrees)
Using a calculator, we find:
d ≈ 13.7 m
Therefore, the pool is approximately 13.7 meters deep.
two children weigh the same. Will this seesaw balance?
yes they would balance because they have the same weight
Explanation:
hope this helps (:
Since the two children have the same weight, they have to sit at equal distances from the pivot of the seesaw. Then they will balance.
A tuning fork is set into vibration above a vertical open tube filled with
water (see figure). The water level is allowed to drop slowly. As it does
so, the air in the tube above the water level is heard to resonate with the
tuning fork when the distance from the tube opening to the water level
is 0.375 and again at 0.625 . No resonance was heard between these
two levels.
(a) What is the frequency of the tuning fork?
(b) What is the minimum value of to hear a resonance?
(c) What is the next level above 0.625 to hear a resonance?
(a) Frequency of the tuning fork is 686 Hz.
(b) Minimum value to hear a resonance is 0.25 m.
(c) The next level above 0.625 to hear a resonance is 0.875 m.
What is meant by resonance frequency ?Resonance frequency is defined as the natural frequency at which the medium vibrates with maximum amplitude.
Here,
(a) Velocity of sound, v = 343 m/s
Distance, L₁ = 0.375 m
L₂ = 0.625 m
From the equation,
ΔL = 1/2 λ
λ = 2ΔL = 2(0.625 - 0.375)
λ = 0.5 m
Therefore,
frequency, f = v/λ
f = 343/0.5
f = 686 Hz
(b) Minimum value to hear a resonance = λ/2 = 0.25 m
(c) Next level above 0.625 to hear resonance,
L₃ = 0.625 + λ/2 = 0.625 +(0.5/2)
L₃ = 0.875 m
Hence,
(a) Frequency of the tuning fork is 686 Hz.
(b) Minimum value to hear a resonance is 0.25 m.
(c) The next level above 0.625 to hear a resonance is 0.875 m.
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Your question was incomplete. Attaching the image here.
the mechanical advantage of a machine is 4 calculate the force required to lift a load of 100 newtons
The force required to lift a load of 100 newtons using machine with mechanical advantage of 4 is 25 newtons.
What is mechanical advantage?The ratio of output force to the input force is called as mechanical advantage of a machine. In this case, mechanical advantage is given as 4.
As, Mechanical Advantage = Output Force / Input Force
4 = Output Force / Input Force
So, Output Force = Mechanical Advantage x Input Force
Output Force = 4 x Input Force
100 newtons = 4 x Input Force
Input Force = 100 newtons / 4
Input Force = 25 newtons
Therefore, the force required to lift a load of 100 newtons using a machine with a mechanical advantage of 4 is 25 newtons.
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If a 40 N block is resting on a rough horizontal table with a coefficient of static friction
If a 40 N block is resting on a rough horizontal table with a coefficient of static friction is 12 N.
What is static friction?Static friction is a force that resists the motion of two objects that are in contact with each other. It is caused by the forces of attraction between the two objects and is usually greater than the force of kinetic friction. The forces of static friction oppose the movement of the two objects and can be overcome by applying a force greater than the static friction.
The maximum force the block can withstand before it starts to move is 40 N multiplied by the coefficient of static friction.
The coefficient of static friction between the block and the table is determined by the materials of the block and the table and the surface roughness of the table.
If the coefficient of static friction is 0.3, then the maximum force the block can withstand before it starts to move is 40 N × 0.3 = 12 N.
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HELP NEEDED ASAP!!!
A crane is lifting a 13,700 N load from the ground to a height of 41.4 m. If this process takes 24.0 seconds, what is the power put out by the crane?
23,600 W
13,700 W
1190 W
570 W
Dividing this by 24.0 s gives a power output of 23,600 W.
What is power?Power is the ability to influence or control the behaviour of people, events, or things. It is the capacity to make decisions and take action in order to effect change. Power can be physical, economic, political, cultural, or social, and can be held by individuals, organizations, or governments. Power can be used positively to create positive change, or negatively to oppress or exploit. It can be used to create a sense of safety, stability, and justice, or to create uncertainty and chaos.
The power output of the crane is calculated by dividing the work done by the crane (the force multiplied by the distance moved) by the time taken to do the work. In this case, the work done is 13,700 N x 41.4 m = 570,380 Nm. Dividing this by 24.0 s gives a power output of 23,600 W.
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A rifle shoots a 4.20 g bullet out of its barrel. The bullet has a muzzle velocity of just as it leaves the barrel. Assuming a constant horizontal acceleration over a distance of 45.0 cm starting from rest, with no friction between the bullet and the barrel, (a) what force does the rifle exert on the bullet while it is in the barrel? (b) Draw a free-body diagram of the bullet (i) while it is in the barrel and (ii) just after it has left the barrel. (c) How many g’s of acceleration does the rifle give this bullet? (d) For how long a time is the bullet in the barrel?
(a)The rifle exerts a force of 628 N on the bullet while it is in the barrel.
(b) (i) There is no friction between the bullet and the barrel, there is no force of friction to consider.
(ii) Just after it has left the barrel, the only force acting on the bullet would be the force of gravity pulling it downward.
(c) The rifle gives the bullet an acceleration of about 24885 g's.
(d) the bullet is in the barrel for 0.030 s.
What is Force?
Force is a physical quantity that describes the interaction between two objects, causing a change in motion. Specifically, it is an influence that can cause an object to accelerate, change direction, or deform. Force is defined as the product of mass and acceleration, according to Newton's second law of motion. Mathematically, force can be represented by the equation:
F = m*a
(a) To determine the force that the rifle exerts on the bullet while it is in the barrel, we need to use the equation for the work done by a constant force:
W = Fd
where W is the work done, F is the force, and d is the distance over which the force is applied. We can rearrange this equation to solve for the force:
F = W/d
We know that the work done on the bullet is equal to the change in kinetic energy:
W = ΔK = (1/2)mv^2
where m is the mass of the bullet and v is its velocity. Plugging in the given values, we get:
W = (1/2)(0.00420 kg)(370 m/s)^2 = 283 J
The distance over which the force is applied is given as 45.0 cm = 0.45 m. So the force exerted by the rifle on the bullet is:
F = W/d = 283 J / 0.45 m = 628 N
Therefore, the rifle exerts a force of 628 N on the bullet while it is in the barrel.
(b) (i) The free-body diagram of the bullet while it is in the barrel would show two forces acting on it: the force of the rifle pushing it forward, and the force of gravity pulling it downward. Since there is no friction between the bullet and the barrel, there is no force of friction to consider.
(ii) Just after it has left the barrel, the only force acting on the bullet would be the force of gravity pulling it downward.
(c) We can calculate the acceleration of the bullet using the formula:
a = Δv/t
where Δv is the change in velocity and t is the time for which the acceleration occurs. We know that the initial velocity of the bullet is 370 m/s and that it starts from rest, so the change in velocity is:
Δv = 370 m/s
The distance over which the acceleration occurs is given as 45.0 cm = 0.45 m. Using the formula for distance traveled with constant acceleration, we can find the time for which the acceleration occurs:
d = (1/2)at^2
0.45 m = (1/2)a(t^2)
Solving for t, we get:
t = sqrt(0.9/a)
Plugging this into the equation for acceleration, we get:
a = Δv/t = 370 m/s / sqrt(0.9/a)
Solving for a, we get:
a = 2.44 x 10^5 m/s^2
To express this acceleration in units of g's, we can divide by the acceleration due to gravity:
a/g = 2.44 x 10^5 m/s^2 / 9.81 m/s^2 = 24885
Therefore, the rifle gives the bullet an acceleration of about 24885 g's.
(d) Using the same formula as before for distance traveled with constant acceleration, we can solve for the time the bullet is in the barrel:
d = (1/2)at^2
0.45 m = (1/2)(2.44 x 10^5 m/s^2)t^2
Solving for t, we get:
t = [tex]\sqrt{(0.00092)}[/tex]s = 0.030 s
Therefore, the bullet is in the barrel for 0.030 s.
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The bullet is in the barrel for 7.05 x 10^-5 seconds. Acceleration can also be caused by changes in direction, such as when an object moves in a circular path.
What is Acceleration?
Acceleration is a fundamental concept in physics and plays an important role in understanding the motion of objects. It is used to describe the behavior of objects in a wide range of applications, from simple everyday situations such as cars accelerating and braking, to more complex phenomena such as the acceleration of particles in particle accelerators or the acceleration of celestial bodies in space.
(a) To find the force exerted by the rifle on the bullet, we can use the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity (muzzle velocity), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance traveled (45.0 cm = 0.45 m).
Rearranging this equation to solve for acceleration:
a = (v^2 - u^2) / 2s
Plugging in the given values:
a = (1200 m/s)^2 / (2 x 0.45 m) = 3.20 x 10^6 m/s^2
The force exerted by the rifle on the bullet can be found using Newton's second law:
F = ma
where F is the force, m is the mass of the bullet (4.20 g = 0.00420 kg), and a is the acceleration we just calculated:
F = 0.00420 kg x 3.20 x 10^6 m/s^2 = 13,440 N
Therefore, the rifle exerts a force of 13,440 N on the bullet while it is in the barrel.
(b)
(i) Free-body diagram of the bullet while it is in the barrel:
The only force acting on the bullet while it is in the barrel is the force exerted by the rifle, which is directed to the right.
|
|
-->| F
|
|
Once the bullet has left the barrel, it is subject to air resistance, which we will assume acts in the opposite direction to the velocity of the bullet. The force of gravity on the bullet is negligible for this problem.
|
|
<--| F_air
|
|
(c) The acceleration given to the bullet can be expressed in terms of g's by dividing by the acceleration due to gravity, g:
a_g = a / g = (3.20 x 10^6 m/s^2) / 9.81 m/s^2 = 326,000 g's
Therefore, the rifle gives the bullet an acceleration of 326,000 g's.
(d) The time the bullet is in the barrel can be found using the kinematic equation:
s = ut + (1/2)at^2
where s is the distance traveled (0.45 m), u is the initial velocity (0 m/s), a is the acceleration we calculated earlier (3.20 x 10^6 m/s^2), and t is the time the bullet is in the barrel (which we want to find).
Rearranging and solving for t:
t = sqrt(2s/a) = sqrt(2 x 0.45 m / 3.20 x 10^6 m/s^2) = 7.05 x 10^-5 s
Therefore, the bullet is in the barrel for 7.05 x 10^-5 seconds.
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Can someone helping me!
For this activity create a kinship chart of a celebrity family or fictional family on television or in other media. Choose a person as “ego” and use the anthropological symbols to identify as many of their relatives as you can. Be sure to choose a family with at least three known generations.
The kinship chart of family is given in the image attached. The kinship chart of the Kardashian-Jenner family with "Kris Jenner" as "ego":
What is the kinship chart?In this kinship chart, Kris Jenner is the "ego" or the main person we are focusing on. Her children are Kourtney Kardashian (daughter) and Robert Kardashian (son). Kourtney has three children: Mason, Penelope, and Reign. Robert has four children: North, Saint, Chicago, and Psalm.
Therefore, Each arrow represents a biological relationship, and each symbol represents a gender and a generational difference. "F" stands for female, "M" stands for male, and "-" stands for unknown gender.
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A soccer ball is kicked off the top of a 12-m high building at 4 m/s. It lands 6.3 m from the base of the building 1.6 seconds later. What is the time?
We can use the kinematic equations to solve for the time it takes for the soccer ball to hit the ground.
Let's use the following variables:
h for the initial height of the soccer ball (12 m)
vi for the initial velocity of the soccer ball (4 m/s)
vf for the final velocity of the soccer ball (0 m/s)
d for the horizontal distance the soccer ball travels before hitting the ground (6.3 m)
t for the time it takes for the soccer ball to hit the ground (what we want to find)
Using the kinematic equation for vertical motion, we can find the time it takes for the soccer ball to reach the ground:
vf^2 = vi^2 + 2*a*h
0 = 4^2 + 2*(-9.8)*12
0 = 16 - 235.2
t_vertical = sqrt(235.2/16) = 3.05 s
So it takes 3.05 seconds for the soccer ball to hit the ground if we consider only the vertical motion.
Next, using the kinematic equation for horizontal motion, we can find the time it takes for the soccer ball to travel the horizontal distance of 6.3 m:
d = vi*t_horizontal
t_horizontal = d/vi = 6.3/4 = 1.575 s
So it takes 1.575 seconds for the soccer ball to travel the horizontal distance of 6.3 m.
Since we know the soccer ball takes the same amount of time to travel horizontally as it does to hit the ground, we can conclude that the time it takes for the soccer ball to hit the ground is 1.575 seconds.
Will sound move faster or slower in water? Why
Answer:
In water, the particles are much closer together, and they can quickly transmit vibration energy from one particle to the next. This means that the sound wave travels over four times faster than it would in air, but it takes a lot of energy to start the vibration.
Explanation:
Answer:
slower as it is more dense in water
Explanation:
Xiomara claims that all the segments in ABC are parallel to the corresponding segments in A'B'C. 1. Write Xiomara's claim as a conjecture. 2. Prove your conjecture. 3. In Xiomara's diagram the scale factor was greater than one. Would your proof have to change if the scale factor was less than one?
1. Conjecture: In triangle ABC and A'B'C', all corresponding segments are parallel if the scale factor is greater than or equal to 1.
How can the conjecture be proven?Let ABC and A'B'C' be two triangles with corresponding sides AB and A'B', BC and B'C', and AC and A'C'. Assume that the scale factor between the two triangles is greater than or equal to 1.
To prove that all corresponding segments are parallel, we need to show that the slopes of these segments are equal.
Consider segment AB and A'B'. Let (x₁, y₁) and (x₂, y₂) be the coordinates of points A and B, and (x₁', y₁') and (x₂', y₂') be the coordinates of points A' and B', respectively. Then, the slope of segment AB is (y₂ - y₁)/(x₂ - x₁), and the slope of segment A'B' is (y₂' - y₁')/(x₂' - x₁').
Since the scale factor is greater than or equal to 1, we have |AB| <= |A'B'|, |BC| <= |B'C'|, and |AC| <= |A'C'|.
Without loss of generality, assume that |AB| <= |A'B'|. Then, we have:
|x₂ - x₁| <= |x₂' - x₁'|, and |y₂ - y₁| <= |y₂' - y₁'|.
Thus, we can conclude that (y₂ - y₁)/(x₂ - x₁) = (y₂' - y₁')/(x₂' - x₁'), which implies that segment AB is parallel to segment A'B'.
Similarly, we can prove that segment BC is parallel to segment B'C' and segment AC is parallel to segment A'C'.
Therefore, we have shown that all corresponding segments are parallel when the scale factor is greater than or equal to 1.
3. If the scale factor is less than one, the proof would need to be modified. In this case, we would have |AB| >= |A'B'|, |BC| >= |B'C'|, and |AC| >= |A'C'|. As a result, we would need to adjust the inequality signs in the proof accordingly. Specifically, we would need to reverse the signs of the inequalities involving the absolute values of the differences between the x and y coordinates of the points in each pair of corresponding segments.
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Newton’s second law says that when an __________________ force is applied to a ________________, it causes it to ____________________________.
Newton’s second law says that when an unbalanced force is applied to a mass, it causes it to accelerate.
What is Newton's law?
Newton's laws are a set of three fundamental principles that describe the behavior of objects in motion, developed by the physicist Sir Isaac Newton in the late 17th century. The laws are as follows:
The first law, also known as the law of inertia, states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity, unless acted upon by an external force.
The second law, also known as the law of acceleration, states that the acceleration of an object is directly proportional to the net force applied to it, and inversely proportional to its mass. This can be expressed mathematically as F = ma, where F is the net force applied, m is the mass of the object, and a is the resulting acceleration.
The third law, also known as the law of action-reaction, states that for every action, there is an equal and opposite reaction. This means that when one object exerts a force on another object, the second object exerts an equal and opposite force back on the first object.
Together, these laws provide a framework for understanding the behavior of objects in motion, and have been applied in countless areas of science and engineering.
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Review questions 1. A fairground ride consists of a large vertical drum that spins so fast that everyone inside it stays pinned against the wall when the floor drops away. The diameter of the drum is 10 m. Assume that the coefficient of static friction between the drum and the rider's clothes is 0.15. b) What is the angular velocity of the drum at this speed?
Angular velocity of the drum at this speed having 15m diameter is 3.6 rad/s
What is Friction ?Friction is a resistance to motion of the object. for example, when a body slides on horizontal surface in positive x direction, it has friction in negative x direction and that measure of friction is a frictional force.
frictional force is directly proportional to the Normal(N).
i.e. [tex]F_{fri}[/tex]∝ N
[tex]F_{fri}[/tex] = μN
where μ is called as coefficient of the friction. It is a dimensionless quantity.
When a body is kept on horizontal surface, its normal will be straight upward which is reaction of mg. i.e. N=mg.
Given,
Diameter of the drum D = 10m , Radius r = 5m
Coefficient of static friction μ = 0.15
To stay everyone pinned against the wall of drum. Frictional Force must be equal to weight mg which are opposite to each other.
μN = mg ........1)
Centrifugal acceleration = Normal
mv²÷r = N
With this equation 1 becomes
μmv²÷r = mg
v² = rg÷μ
v² = 5m*9.8m/s ÷ 0.15
v² = 326.6
v= = 18.07 ~ 18 m/s
Hence minimum linear velocity required for the drum is 18m/s.
for angular velocity of the drum, V=rω
ω = v÷r
ω = 18m/s ÷ 5m
ω = 3.6 rad/s
Hence angular velocity of the drum at 18m/s speed is 3.6 rad/s
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1. A 2 kg object traveling at 5 m/s collides with a stationary 3 kg object. What is the final velocity of the two objects if they stick together after the collision?
2. A 1000 kg car is moving at 10 m/s when it collides with a stationary 500 kg car. If the two cars stick together after the collision, what is their final velocity?
3. A 50 kg boy is riding a 10 kg bike at 5 m/s. The boy jumps off the bike and lands on the ground. What is the velocity of the bike after the boy jumps off?
4. Two hockey players, one with a mass of 75 kg and the other with a mass of 85 kg, collide head-on while skating towards each other at 8 m/s. If they stick together after the collision, what is their final velocity?
5. A 2 kg ball is rolling at 5 m/s when it collides with a stationary 1 kg ball. After the collision, the 2 kg ball is moving at 2 m/s. What is the final velocity of the 1 kg ball?
6. A 500 kg rocket is traveling at 500 m/s. It releases a 50 kg satellite at a speed of 1000 m/s in the opposite direction. What is the velocity of the rocket after the satellite is released?
7. A 5 kg mass moving at 10 m/s collides with a stationary 2 kg mass. After the collision, the 2 kg mass moves away at 4 m/s. What is the final velocity of the 5 kg mass?
8. A 1 kg mass moving at 2 m/s collides with a stationary 2 kg mass. After the collision, the 1 kg mass moves away at 1 m/s. What is the final velocity of the 2 kg mass?
Answer:
1. To solve this problem using conservation of momentum, we can use the equation:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf
where m1 and v1 are the mass and velocity of the first object, m2 and v2 are the mass and velocity of the second object, and vf is the final velocity of the two objects after they stick together.
Plugging in the given values, we get:
(2 kg * 5 m/s) + (3 kg * 0 m/s) = (2 kg + 3 kg) * vf
Simplifying the equation, we get:
10 kg m/s = 5 kg * vf
vf = 2 m/s
Therefore, the final velocity of the two objects after the collision is 2 m/s.
2. Using the same equation as above and plugging in the given values, we get:
(1000 kg * 10 m/s) + (500 kg * 0 m/s) = (1000 kg + 500 kg) * vf
Simplifying the equation, we get:
10,000 kg m/s = 1500 kg * vf
vf = 6.67 m/s
Therefore, the final velocity of the two cars after the collision is 6.67 m/s.
3. Before the boy jumps off the bike, the total momentum of the system is:
(50 kg + 10 kg) * 5 m/s = 300 kg m/s
After the boy jumps off, the total momentum of the system is:
50 kg * v_bike
Using conservation of momentum, we can equate the two and solve for v_bike:
(50 kg + 10 kg) * 5 m/s = 50 kg * v_bike
Simplifying the equation, we get:
300 kg m/s = 50 kg * v_bike
v_bike = 6 m/s
Therefore, the velocity of the bike after the boy jumps off is 6 m/s.
4. Using the same equation as in problem 1 and plugging in the given values, we get:
(75 kg * 8 m/s) + (85 kg * -8 m/s) = (75 kg + 85 kg) * vf
Simplifying the equation, we get:
0 = 1600 kg * vf
vf = 0 m/s
Therefore, the final velocity of the two players after the collision is 0 m/s.
5. Using conservation of momentum and plugging in the given values, we can set up the equation:
(2 kg * 5 m/s) + (1 kg * 0 m/s) = (2 kg + 1 kg) * 2 m/s + (1 kg * vf)
Simplifying the equation, we get:
10 kg m/s = 5 kg * 2 m/s + vf
vf = 0 m/s
Therefore, the final velocity of the 1 kg ball after the collision is 0 m/s.
6. To solve this problem, we can use the conservation of momentum equation again:
(500 kg * 500 m/s) = (450 kg * vf_r) + (50 kg * 1000 m/s)
where vf_r is the final velocity of the rocket after releasing the satellite.
Simplifying the equation, we get:
250,000 kg m/s = 450 kg * vf_r
vf_r = 555.56 m/s
Therefore, the velocity of the rocket after releasing the satellite is 555.56 m/s.
7. Using the same conservation of momentum equation and plugging in the given values, we get:
(5 kg * 10 m/s) + (2 kg * 0 m/s) = (5 kg + 2 kg) * vf
Simplifying the equation, we get:
50 kg m/s = 7 kg * vf
vf = 7.14 m/s
Therefore, the final velocity of the 5 kg mass after the collision is 7.14 m/s.
8. Using the same conservation of momentum equation and plugging in the given values, we get:
(1 kg * 2 m/s) + (2 kg * 0 m/s) = (1 kg + 2 kg) * 1 m/s + (2 kg * vf)
Simplifying the equation, we get:
2 kg m/s = 3 kg + (2 kg * vf)
vf = -0.5 m/s
Therefore, the final velocity of the 2 kg mass after the collision is -0.5 m/s. This negative velocity indicates that the mass is moving in the opposite direction to the initial direction.
According to the solving the final velocity of the two objects after they stick together is 1.67 m/s.
What do you mean by velocity?The velocity of a body determines whether it is moving away from the ground or towards an object. Speed is often a scalar quantity. Velocity is a vector quantity in its most basic form. It gauges the rate of change in a distance. It relates to how quickly displacement is altering.
According to the given information:the objects stick together after the collision, we can use conservation of momentum to solve for the final velocity of the combined mass:
m1 * v1 + m2 * v2 = (m1 + m2) * vf
where m1 and v1 are the mass and velocity of the first object, m2 and v2 are the mass and velocity of the second object, and vf is the final velocity of the combined mass.
Substituting the given values:
(2 kg) * (5 m/s) + (3 kg) * (0 m/s) = (2 kg + 3 kg) * vf
Simplifying and solving for vf:
vf = 1.67 m/s
the final velocity of the two objects after they stick together is 1.67 m/s.
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I NEED HELP PLEASE !!!!
The average speed of the skier is 53.3 m/min.
option C.
What is the average speed of the skier?Average speed is a measure of how fast an object or a person travel over a certain distance in a specific amount of time.
It is calculated by dividing the total distance traveled by the time it took to travel that distance.
The formula for average speed is:
Average Speed = Total Distance Traveled / Time Taken
The average speed of the skier is calculated as follows;
Average speed = ( 160 m ) / 3 mins
Average speed = 53.3 m/min
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URGENT HELP NEEDED!!!
Two spheres of molten rock, both moving in the positive y-direction, collide in space. One sphere has a mass of 660 kg and is moving at a velocity of 17 m/s. The other sphere has a mass of 970 kg, and is moving at a velocity of 24 m/s. When the two spheres collide, and become one unit, what will be their combined velocity?
21.2 m/s
24.0 m/s
41.0 m/s
20.5 m/s
Answer:
20.5 m/s
Explanation:
There are three stable atoms of Argon (Atomic Number 18): Argon-36, Argon-38 and Argon-40. What would the atoms of these isotopes have in common? What would be different about their atoms?
please answer ASAP. It's for a test.
While the three stable isotopes of Argon share the same number of protons and electrons, they differ in their atomic masses due to varying numbers of neutrons.
This means they have identical electron configurations and chemical properties. Additionally, they all have the same number of electrons, 18, which determines their chemical behavior and bonding capabilities.
However, the isotopes differ in their neutron numbers, which gives rise to their atomic masses. Argon-36 has 18 protons and 18 neutrons, Argon-38 has 18 protons and 20 neutrons, and Argon-40 has 18 protons and 22 neutrons.
This variation in neutron count leads to differences in atomic mass. Consequently, the isotopes exhibit different atomic weights, with Argon-36 having a mass of approximately 36 atomic mass units (amu), Argon-38 around 38 amu, and Argon-40 approximately 40 amu.The differing atomic masses affect the isotopes' physical properties.
For instance, the isotopes may have different boiling points, melting points, or densities due to variations in the average mass of their atoms. These subtle differences are relevant in scientific research, especially in fields like geochronology and radiometric dating, where the abundance ratios of these isotopes are utilized to determine the age of rocks and minerals.
In summary, These differences give rise to variations in their atomic weights and physical properties.
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Cart 1 is stationary and has the same mass of cart 2 which collides with cart 1 at a speed of 10 m/s. They do not stick together. Cart 2's velocity after the collision is
a. 5 m/s
b. 20 m/s
c. 0.0 m/s
d. 10 m/s
What is the power involved in lifting a 20- kg object 1.0m in 1.0 s?
Answer:
Explanation:
Power=work done/time
work done=force* displacement
force=mass*acceleration due to gravity
Therefore, power= mass*acceleration due to gravity*displacement/time
20*9.8*1/1=196watts
There are three solids made up of aluminium steel and wood of the same shape and same volume which of them would have highest interia
The interia is varied and it depends on the mass, itt will be sameshape and size so it will be light closer.
What is volume ?
Space is occupied by every three-dimensional object. Its volume serves as a gauge for this area. The area contained by an object's limits in three-dimensional space is referred to as its volume. Another name for it is an object's capacity.
What is element ?
The atoms make the element. The atoms are created up of the element; it is a pure substance it can't be able to split up. The element is created up of a single kind of atom. Atoms are the main thing for the creation of elements. Elements are classified according to their performance and physical properties. Elements are placed based on the performance in the Mendeleev periodic table.
Therefore, The interia is varied and it depends on the mass, itt will be sameshape and size so it will be light closer.
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M
Dopamine is key for
Functioning of autonomic system.
Release of glucose
✔Pleasure
Reinforcement
Blood pressure
Dopamine is key for : pleasure. Dopamine lets you feel pleasure, satisfaction and also motivation. When one feels good that they have achieved something, then it is because one has surge of dopamine in the brain.
What is the role of dopamine?Dopamine is key for pleasure, reinforcement, and motivation and is commonly known as the "reward molecule". It is responsible for the feelings of pleasure and satisfaction that we experience after completing rewarding activity, like eating food or engaging in social interaction.
Dopamine is also involved in reinforcement, which is the process by which brains learn to associate specific actions or behaviors with rewards.
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1. If the first harmonic of a guitar string is 568 Hz, what is the frequency of the fifth harmonic?
Answer:
Look down below
Explanation:
300Hz is the 5th harmonic in a 60 Hz system, or the 6th harmonic in a 50 Hz system.
Could someone help me with the pre-lab questions and listing the Independent, dependent, and controlled variables please?
Pre-lab Questions:
Describe the kinetic molecular theory.
What is thermal energy and conduction?
Explain the differences between open, closed, and isolated systems.
Hypothesis
Before you begin your lab, review the procedures. Then write a hypothesis that reflects what you think will happen to the water temperature. Record your hypothesis as an “if, then” statement.
Variables
List the independent, dependent, and controlled variables.
Materials
Hot water
Cold water
Small, plastic or glass cups
Thermometer
Measuring cups
Procedure
Gather materials.
Measure 50 ml of hot water and 50 ml of cold water into separate cups.
Use a thermometer to measure the temperature of hot water and the cold water. Record your prediction for the final temperatures.
Pour the hot and cold water into a third cup.
Allow some time for the thermometer to reach a final temperature, then measure the temperature of the water. Record the temperature in the data table.
Repeat for a total of three trials.
If I mix hot and cold water together, then the final temperature of the water will be somewhere in between the initial temperatures of the hot and cold water.
What is temperature?Temperature is a measure of heat energy, which is defined as the average kinetic energy of particles in a substance. It is measured using a thermometer, and is expressed in either Celsius (°C) or Fahrenheit (°F). Temperature is an important physical property as it affects the rate of chemical reactions, the speed of sound and light, and the behavior of matter.
It is also used to measure the intensity of heat energy in terms of the amount of thermal energy generated or absorbed by an object or material. Temperature is a fundamental physical property and is used to quantify the energy of a system. Temperature increases when energy is added to a system, and decreases when energy is removed.
Hypothesis:If I mix hot and cold water together, then the final temperature of the water will be somewhere in between the initial temperatures of the hot and cold water.
Independent Variable: Temperature of hot and cold water.
Dependent Variable: Final temperature of the water.
Controlled Variables:
Amount of hot and cold water, type of cups, thermometer, and measuring cups.
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Which of the following candles would most likely melt the slowest on a sunny day and why?
Question 2 options:
A red candle
A black candle
A blue candle
A white candle
Because it absorbs all the colors of light
Because it reflects all the colors of light
Because it refracts all the colors of light
Because it transmits all the colors of light
Answer:White Candle
Explanation:because it reflects all colors of light.
Q12. Calculate the resistance of a microwave with 5V and a current of 300mA
Answer:
To calculate the resistance of the microwave, we can use Ohm's Law, which states that:
resistance = voltage / current
Substituting the given values into this equation, we get:
resistance = 5V / 0.3A
resistance = 16.67 ohms
Therefore, the resistance of the microwave with 5V and a current of 300mA is 16.67 ohms.
. 4. What is the trend for your data (pay attention to the number of pieces you started with compared to the number of pieces you ended up with). How does this data represent the Law of Conservation of Mass? Add Content
We can see here that the trend of the data shows in each trial, the number of bread and cheese used drastically reduced or are used up after each reaction.
The data represents the Law of Conservation of Mass by showing that if you add two elements you will have one product.
What is the Law of Conservation of Mass?The Law of Conservation of Mass is a fundamental principle in chemistry, which states that the total mass of the reactants in a chemical reaction is equal to the total mass of the products.
In other words, the total mass of matter in a closed system remains constant during a chemical reaction, and no matter is created or destroyed.
The part that completes the question is seen below:
9. Conduct the SAME trials as you did above, but this time count the number of ingredients you started with and what you ended up with. You are NOT counting sandwiches, but all the individual ingredients.
Trial Number 1 2 3 4 5
#of Bread Before Reaction 5 6 3 5 6
#of Cheese Before Reaction 3 4 4 2 5
#of Bread After Reaction 1 0 1 1 0
#of Cheese After Reaction 1 1 3 0 2
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CAN SOMEBODY HELP ME WITH ALL THIS PLEASE
Electromagnetic induction is the process of generating an electric current by moving a conductor through a magnetic field.
What is electromagnetic induction?As you move the magnet in and out of the coil, it induces a current in the coil, which generates an electrical signal. This is known as electromagnetic induction, and the strength of the induced current depends on the strength and speed of the magnetic field.Learn more about electromagnetic induction, here:
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A car accelerates uniformly from rest to a speed of 40.0 mi/h in 13.0 s.
(a) Find the distance the car travels during this time.
m
(b) Find the constant acceleration of the car.
m/s2
Answer:
b
Explanation:
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