(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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# Calculate the radius of water drop which look just remain suspendat in an electric-ff- eld of 300 viem and charged with one election-
The radius of a water droplet suspended in an electric field can be calculated using the following formula:
r = (3qE/4πρg)^(1/2)
where r is the radius of the droplet, q is the charge on the droplet, E is the strength of the electric field, ρ is the density of the droplet material (assumed to be that of water, which is 1000 kg/m^3), and g is the acceleration due to gravity.
What is the radius of water drop which look just remain suspendat in an electric-ff- eld of 300 viem and charged with one election?In this case, q = -1.602 × 10^-19 C (the charge on an electron), E = 300 V/m, ρ = 1000 kg/m^3, and g = 9.81 m/s^2.
Plugging in these values, we get:
r = (3(-1.602 × 10^-19 C)(300 V/m)/(4π(1000 kg/m^3)(9.81 m/s^2)))^(1/2)
= 1.83 × 10^-5 m
Therefore, the radius of the water droplet is approximately 1.83 × 10^-5 meters.
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a rock rolling down a slope from rest covers a distance of 4 m in the first second. What distance will it covers in 3 sec?
Answer: 12 meters
Explanation:
the rate is 4 meters per second
the rock rolled for 3 seconds
4 x 3 = 12
A 500 g object is dropped from a height of 2 meters. What is its kinetic energy just before it hits the ground?
Answer: 9.8 J
Explanation:
Since the gravitational potential energy of the object is mgh or mass*acceleration due to gravity*initial height, its [tex]U_{g}[/tex] is 9.8 J. Due to the Law of Conservation of Energy, its kinetic energy will also be 9.8 J. This can be seen in the equation [tex]KE_{i}+ PE_{i}= KE_{f} + PE_{f}[/tex]. Since there is no initial kinetic energy and no final potential energy, its initial potential energy is equal to its final kinetic energy.
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 mass is vibrating on a string. its frequency describes ( )
Answer:
The frequency of a mass vibrating on a string describes the number of complete cycles of vibration that occur per unit of time, usually measured in hertz (Hz).
Explanation:
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:
あなたは本当に知りたいですか？私はあなたが答えを理解する方法ではなく、答えのためにここにいるという意味ですので、答えを書き留めて正しいことができますので、あなたの幸せを願っています
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:
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.
I = 1/2MR^2 is rolling with a translational speed v along a horizontal surface without slipping. The work required to stop the disk is
A. 1/2 mv^2
B. 1/2mv^2 + 1/2Iw^2
C. 1/2Iw^2
D. 1/2mv^2 - 1/2Iw^2
Answer:
B. 1/2mv^2 + 1/2Iw^2
Explanation:
Ok so my question is “ blank heat is a measure of how much energy it takes to raise the temperature of a substance.” Help PLEASEE I’m stuck!
Specific heat is a measure of how much energy it takes to raise the temperature of a substance.
What is temperature ?Temperature is defined as the measurement of degree of amount of hotness or coldness of a body.
Here,
Specific heat is a measure of how much energy it takes to raise the temperature of a substance. More clearly, specific heat is the amount heat required to raise the temperature of unit mass through one degree.
If an amount of heat Q is given to a body of mass m and ΔT is the rise in temperature. Then specific heat capacity,
C = Q/mΔT
Its unit is Jkg⁻¹K⁻¹
Specific heat of a substance is a constant but it changes slightly with change in temperature.
The rise in temperature is small for body having large specific heat.
Hence,
Specific heat is a measure of how much energy it takes to raise the temperature of a substance.
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If the energy released by an electron making a transition from one hydrogen atom orbit to another is 3.02 ✕ 10^−19 J, what is the wavelength (in nm) of the photon?
Answer:
The wavelength of the photon can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 ✕ 10^−34 J s), c is the speed of light (2.998 ✕ 10^8 m/s), and λ is the wavelength.
λ = hc/E = (6.626 ✕ 10^−34 J s) (2.998 ✕ 10^8 m/s) / (3.02 ✕ 10^−19 J) = 656.4 nm
Explanation:
A hockey puck with a mass of 0.171 kg slides over the ice. The puck initially slides with a speed of 5.65 m/s, but it comes to a rough patch in the ice which slows it down to a speed of 3.05 m/s.
How much energy is dissipated as the puck slides over the rough patch?
To answer the question we have, 1.93401 J of energy is lost when the puck travels over the uneven surface.
What is the easiest way to define energy?Energy is referred to by scientists as the capacity for work. People have figured out how to transform energy from one kind to the other before employing it to accomplish tasks, making western civilisation possible.
Hockey weighs 0.171 kg.
Starting speed: 5.65 m/s
Ultimately, the speed was 3.05 m/s.
We must determine how much kinetic energy was lost on the tough patch.
[tex]E_{d}=\frac{1}{2} v^2_{2} -v^2_1[/tex]
Where, m = mass
v₁ = Initial velocity
v₂ = Final velocity
Put the value into the formula
[tex]E_{d}=\frac{1}{2}[/tex] × 0.171 × [tex](3.05^2 - 5.65^2)[/tex]
= 1.93401 J energy is dissipated as the puck slides over the rough patch.
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What type of wave is:
a) Light
b) Sound
Answer:
Explanation:
a) transverse
b)longitudinal waves
Assume that bone will fracture if a shear stress more than 9.00 × 10^7 N/m^2 is exerted. What is the maximum force that can be exerted on the femur bone in the leg if it has a minimum effective diameter of 9.00 cm ? Young’s modulus for bone is 1.8 × 10^10 N/m^2 Answer in units of kN
The area of the bone is 0.785 m². Then the maximum force that can be exerted on the femur bone if the shear stress is 9 × 10⁷ N/m² is 70650 kN.
What is young's modulus ?The Young's modulus of a material is the ratio of its stress to strain. Where stress is the force per unit area and strain be the ratio of change in length to the original length.
given stress s = 9 × 10⁷ N/m²
diameter of the bone d = 9 cm = 0.09 m.
then area = π d²/4
a = 3.14 × (0.09 m )²/4 = 0.785 m².
Stress = maximum force/area
then Fmax = stress × area
Fmax = 9 × 10⁷ N/m² × 0.785 m²
= 70650 kN.
Therefore, the maximum force that can be exerted to the bone is 70650 kN.
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Kyle is blowing leaves with a leaf blower. He lifts the 1600 newton leaf blower at a
distance of 6 meters. What are the joules of work being put out?
PLS HELP <3
Answer:
To calculate the work being done, we need to use the formula:
work = force x distance x cos(theta)
where force is in newtons, distance is in meters, and theta is the angle between the direction of the force and the direction of the movement.
In this case, the force is the weight of the leaf blower, which is 1600 N, the distance is 6 meters, and the angle between the force and the movement is 0 degrees (since Kyle is lifting the leaf blower straight up). So we have:
work = 1600 N x 6 m x cos(0°)
work = 9600 J
Therefore, the joules of work being put out by Kyle are 9600 J.
Explanation:
When weight training you should aim to be able to lift the weight ____ to _____ times without strain or injury but with enough resistance to give the muscles a good workout. 1 to 3 30 to 40 8 to 12 Show
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:
A person can jump a horizontal distance of 1.31 m on the Earth. The acceleration of gravity is 12.3 m/s^2.
a) How far could he jump on the Moon, where the free-fall acceleration is 0.258g Answer in units of m.
b) Repeat for Mars, where the acceleration due to gravity is 0.293g. Answer in units of m.
Answer:
a) On the Moon, where the acceleration due to gravity is 0.258g:
First, we need to find the acceleration due to gravity on the Moon:
g_Moon = 0.258g_Earth
g_Moon = 0.258(12.3 m/s^2)
g_Moon = 3.17 m/s^2
Now we can use the range formula for projectile motion to find the distance he could jump:
R = (v^2/g) sin(2θ)
Assuming the same initial velocity and angle of jump, we can rearrange the formula to solve for R:
R = (v^2/g) sin(^2/g_Earth) sin(2θ) * (g_Moon/g_Earth)
R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.17 m/s^2) / (12.3 m/s^2)
R = 0.191 m
Therefore, he could jump approximately 0.191 m on the Moon.
!
a) On the Moon, where the acceleration due to gravity is 0.258g:
First, we need to find the acceleration due to gravity on the Moon:
g_Moon = 0.258g_Earth
g_Moon = 0.258(12.3 m/s^2)
g_Moon = 3.17 m/s^2
Now we can use the range formula for projectile motion to find the distance he could jump:
R = (v^2/g) sin(2θ)
Assuming the same initial velocity and angle of jump, we can rearrange the formula to solve for R:
R = (v^2/g) sin(
network error
^2/g_Earth) sin(2θ) * (g_Moon/g_Earth)
R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.17 m/s^2) / (12.3 m/s^2)
R = 0.191 m
Therefore, he could jump approximately 0.191 m on the Moon.
b) On Mars, where the acceleration due to gravity is 0.293g:
Similarly, we need to find the acceleration due to gravity on Mars:
g_Mars = 0.293g_Earth
g_Mars = 0.293(12.3 m/s^2)
g_Mars = 3.61 m/s^2
Using the same formula and rearrangement as in part a, we can find the distance he could jump on Mars:
R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.61 m/s^2) / (12.3 m/s^2)
R = 0.223 m
Therefore, he could jump approximately 0.223 m on Mars.
If you stretched a rubber band so that it had 100-J of potential energy, with how much kinetic energy will the rock leave the slingshot?
Answer:
less than 100-J
Explanation:
The potential energy held in the stretched rubber band is turned into kinetic energy of the rock when it is released, assuming that the rubber band is used to launch a rock from a slingshot.
The mass of the rock and the effectiveness of the slingshot in transmitting the energy from the rubber band to the rock are two elements that affect how much kinetic energy the rock will have. To estimate the kinetic energy, though, we may make certain generalizations.
Assume that no energy is lost as a result of friction or air resistance and that the entire potential energy held in the rubber band is transformed into the kinetic energy of the rock. In this hypothetical situation, the potential energy of the stretched rubber band would be equal to the kinetic energy of the rock.
As a result, the rock will have 100 J of kinetic energy when it exits the slingshot if the rubber band contains 100 J of potential energy. The actual kinetic energy of the rock would be less than 100 J since some energy will be wasted owing to things like friction.
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What is the wavelength (in nm) of maximum intensity for a celestial body with a temperature of 50000 K?
The wavelength of maximum intensity for a celestial body with a temperature of 50000 K is approximately 577 nm.
What is wavelength?To find the wavelength of maximum intensity for a celestial body with a temperature of 50000 K, we can use Wien's displacement law, which relates the temperature of an object to the wavelength at which it emits the most radiation. The law is given by:
λ_max = b / T
where λ_max is the wavelength of maximum intensity, b is Wien's displacement constant (2.898 × 10⁻³ m·K), and T is the temperature in Kelvin.
We can convert the temperature of 50000 K to Kelvin by adding 273.15 to get:
T = 50000 K + 273.15 = 50273.15 K
Plugging in the values, we get:
λ_max = (2.898 × 10⁻³ m·K) / 50273.15 K
Simplifying, we get:
λ_max = 5.77 × 10⁻⁸ meters
To convert meters to nanometers, we can multiply by 10⁹:
λ_max = 5.77 × 10⁻⁸ meters × 10⁹ nm/m = 577 nm
Therefore, the wavelength of maximum intensity for a celestial body with a temperature of 50000 K is approximately 577 nm.
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Can someone please teach me how to do this
I need to find the tension in the bottom rope
The solutions for questions A and B are
28.1866N T_1=0 NWhat is tension?Generally, Tension is a physical force that pulls on an object, often tending to stretch it. It is typically measured in units of force per area, such as newtons per square meter (N/m2). Tension is an important concept in mechanics, physics, engineering, and other fields.
(a) T_1=m_1(g-a)
T_1=4.81 (9.8-3.94)
T_1=28.1866N
(b) In freefall
a=g
T_1=m_1(g-g)
T_1=0 N
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Urgent!!!
Two students are testing out the law of conservation of momentum, by throwing balls of clay so that they collide. One student throws a 535g ball north at 12.4 m/s. The other throws a 725g ball south at 6.4 m/s. When the two balls of clay collide and form a single unit, what is the velocity of the combined unit?
1.86 m/s south
8.74 m/s north
1.86 m/s north
8.74 m/s south
Answer:
To solve this problem, we need to use the law of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision.
The momentum of an object is given by its mass multiplied by its velocity, so we can calculate the initial momentum of each ball before the collision:
- The northbound ball has a momentum of 535g * 12.4 m/s = 6644 g*m/s north
- The southbound ball has a momentum of 725g * (-6.4 m/s) = -4640 g*m/s north (note that the negative sign indicates southward direction)
Adding these momenta together, we get a total momentum of 6644 g*m/s - 4640 g*m/s = 2004 g*m/s north.
After the collision, the two clay balls stick together and move as a single unit. Let's call the mass of the combined unit "M" and its velocity "v". By conservation of momentum, we know that the total momentum of the combined unit after the collision must be the same as the total momentum before:
M * v = 2004 g*m/s north
To solve for v, we need to figure out the mass of the combined unit. This is simply the sum of the masses of the two original balls:
M = 535g + 725g = 1260g
Substituting this into the equation above, we get:
1260g * v = 2004 g*m/s north
Solving for v, we get:
v = 1.59 m/s north
Therefore, the combined unit moves 1.59 m/s north after the collision.
However, the answer choices given in the problem are in meters per second, not meters per second north/south. To convert the answer, we need to add a direction. Recall that the northbound ball had a positive velocity and the southbound ball had a negative velocity. Since the combined unit is moving northward, we know its velocity must be positive.
Therefore, the final answer is 1.59 m/s north, which corresponds to answer choice C.
I can’t figure it out
Answer:
Explanation: In anyway can you get a better picture just a close picture i cant read it
What is the velocity of a 1,000.0 kg car if its kinetic energy is 200 kJ?
Answer: 20
Explanation:
3. A shaft of 100 mm diameter rotates at 120 rad/s in a bearing 150 mm long. If the dial clearance is 0-2 mm and the absolute viscosity of the lubricant is 20 kg/ms find the power loss in the bearing.
The power loss in the bearing is 1.36 W shaft of 100 mm diameter rotates at 120 rad/s in a bearing 150 mm long.
What is viscosity?The viscosity of a fluid is a measure of its resistance to flow. A fluid with a high viscosity resists movement because its molecular structure causes a lot of internal friction.
The viscosity of a fluid is a measure of its resistance to flow. A fluid with a high viscosity resists movement because its molecular structure causes a lot of internal friction.
To calculate the power loss in the bearing, we can use the formula:
P = F × V
To calculate the frictional force, we can use the formula:
F = μ × A × P
Where μ is the coefficient of friction, A is the area of the bearing, and P is the pressure of the lubricant.
First, we need to calculate the pressure of the lubricant:
P = μ × Viscosity × (V/D)
Where D is the diameter of the shaft.
P = (0.02) × 20 × (120/0.1) = 480 N/[tex]m^2[/tex]
Next, we can calculate the area of the bearing:
A = π × (D/2)^2 × L
Where L is the length of the bearing.
A = π × (0.1/2)^2 × 0.15 = 0.001178 [tex]m^2[/tex]
Now we can calculate the frictional force:
F = μ × A × P = (0.02) × 0.001178 × 480 = 0.0113 N
Finally, we can calculate the power loss:
P = F × V = 0.0113 × 120 = 1.36 W
Therefore, the power loss in the bearing is 1.36 W.
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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.
a) What is the minimum speed required for the riders so that
they stay pinned against the inside of the drum when the
floor drops away?
b) What is the angular velocity of the drum at this speed?
a) Minimum speed required for the riders so that they stay pinned against the inside of the drum is 18m/s.
b) Angular velocity of the drum at this speed having diameter of the drum is 10 m 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 = 0.15
a) 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
μ[tex]\frac{mv^{2} }{r}[/tex] = mg
v² = rg÷μ
v² = 5m*9.8m/s ÷ 0.15
v² = 326.6
v= [tex]\sqrt{326.6}[/tex] = 18.07 ~ 18 m/s
Hence minimum speed required for the riders is 18m/s.
b) 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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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
a charged cloud system produces an electric field in the air near the earth surface. When a particle (q=-2.0x10-9)is acted on by a downward electrostatic force of 3.0x10^-6 N when placed in this field, determine the magnitude of the electric field
The magnitude of the electrical field would be 1.5x10^3 N/C.
Electrical field calculationThe electrostatic force experienced by a charged particle in an electric field is given by the formula:
F = qE
Where F is the electrostatic force, q is the charge of the particle, and E is the electric field strength.
In this case, the electrostatic force acting on the particle is 3.0x10^-6 N and the charge of the particle is -2.0x10^-9 C. So we have:
3.0x10^-6 N = (-2.0x10^-9 C)E
Solving for E, we get:
E = (3.0x10^-6 N) / (-2.0x10^-9 C)
E = -1.5x10^3 N/C
Since the electric field is a vector quantity, its magnitude is always positive. Therefore, the magnitude of the electric field in this case is:
|E| = 1.5x10^3 N/C
So the magnitude of the electric field is 1.5x10^3 N/C, directed downward.
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As we've discussed, one model of material deformation links stress and strain by a material constant called a modulus. For a material under lengthwise stress (tension or compression), this is Young’s modulus (Y) and the relation may be stated as:
The data below was collected for five different materials using tools which allowed for the direct measurement of the stress and strain. By applying the model above, rank the Young’s modulus for each of these five materials from largest to smallest.
Young’s modulus for each of these five materials from largest to smallest are mentioned below.
What is Young’s modulus?
Many substances lack linearity and elasticity after very minor deformation. Only materials that are linearly elastic are subject to the constant Young's modulus. In this case, the ratio of stress to strain, which corresponds to the material's stress, determines the Young's modulus.
What is data?
Facts such as numbers, words, measurements, observations, or even simple descriptions of objects are grouped together as data. Both qualitative and quantitative data are possible. Qualitative data is information that is descriptive (describes something), whereas discrete data can only take particular values (like whole numbers).
Therefore, Young’s modulus for each of these five materials from largest to smallest are mentioned above.
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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: