The kinetic energy (KE) of an object is given by the formula:
KE = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
Plugging in the given values, we get:
KE = (1/2) * 60 kg * (20 m/s)^2
KE = 12,000 joules
Therefore, the kinetic energy of the car is 12,000 joules.
a hockey puck with mass 0.170 kg traveling along the blue line at 1.50 m/s strikes a stationary puck with the same mass. The first puck exits the collision in direction that is 30 degrees away from the blue line at a speed of 0.75 m/s .What is the direction and magnitude of the velocity of the second puck after the collision? Is this an elastic collision?
The KEf is less than KEi, we can conclude that this is not an elastic collision. Some kinetic energy was lost during the collision, possibly due to friction between the pucks or deformation of the pucks. The final velocity of the second puck is also 0.382 m/s, since the two pucks stick together after the collision.
What is the conservation of kinetic energy?
The conservation of kinetic energy is a fundamental principle in physics that states that in an isolated system, the total kinetic energy of the system remains constant, provided that no external forces act on the system. This means that the kinetic energy of the system is conserved during any process or interaction between the objects in the system. In other words, the total kinetic energy of the system before and after the interaction is equal.
We can use momentum conservation and kinetic energy conservation to solve this problem.
First, let's find the initial momentum of the system. Since the second puck is stationary, its momentum is zero. The momentum of the first puck is:
p1i = m1v1i = (0.170 kg)(1.50 m/s) = 0.255 kg m/s
The system's initial total momentum is:
pTotali = p1i + p2i = 0.255 kg m/s
After the collision, the first puck is moving in a direction that is 30 degrees away from the blue line at a speed of 0.75 m/s. Let's break this velocity into its x and y components:
vx1f = v1f cos(30) = 0.75 m/s cos(30) = 0.65 m/s
vy1f = v1f sin(30) = 0.75 m/s sin(30) = 0.375 m/s
Now, let's use conservation of momentum to find the final velocity of the second puck. Since the collision is inelastic, the two pucks will stick together after the collision, so their final velocities will be the same.
pTotalf = pTotali
(m1 + m2)vf = m1v1f + m2v2f
Replacing the values we know, we get:
(0.34 kg)vf = (0.170 kg)(0.65 m/s) + (0.170 kg)(0 m/s)
Solving for vf, we get:
vf = 0.382 m/s
The final velocity of the second puck is also 0.382 m/s, since the two pucks stick together after the collision.
To check whether this is an elastic collision, we can use the conservation of kinetic energy. The system's initial kinetic energy is:
(1/2)m1v1i2 + (1/2)m2(0 m/s)2 = 0.0574 J
The system's total kinetic energy is:
KEf = (1/2)(m1 + m2)vf² = 0.0277 J
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For question 11: the cutoff on the top right says "mercury but not as deeply as before."
When the rest of a half-filled container containing a steel ball floating on the surface of the mercury is filled with water, the ball remains partially submerged in mercury but not as deeply as before.
How does the density of a substance determine whether it floats or sinks in a liquid?Whether a substance will float or sink in a liquid is determined by its density relative to the density of the liquid.
If the density of the substance is less than the density of the liquid, it will float because it is less dense than the liquid and will experience a buoyant force that is greater than its own weight. If the density of the substance is greater than the density of the liquid, it will sink because it is denser than the liquid and will experience a buoyant force that is less than its own weight.
For example, water will float on mercury since it is less dense than mercury.
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Complete question:
Consider a steel ball floating on the surface of the mercury in a half-filled container.
What happens when the rest of the container is filled with water? Mercury is denser than steel, and both are denser than water. Mercury and water do not mix.
1. The ball remains partially submerged in mercury to exactly the same depth as before.
2. The ball Boats to the surface of the water.
3. Mercury floats on top of the water, and the ball floats on the mercury's surface
4. The ball remains partially submerged in mercury but not as deeply as before.
5. The ball sinks to the bottom of the container
6. The ball remains partially submerged in mercury to a greater depth than before.
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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How is the plot of output of dc generator different from the plot of output for an ac generator
The key difference between the two types of generators is that DC generators produce a constant and unidirectional output while AC generators produce a varying and alternating output.
How does the AC and DC generator works ?The output of a DC generator produces a steady, unidirectional current that flows in one direction. The plot of its output voltage versus time would be a straight line with no fluctuations or changes in direction.
On the other hand, an AC generator produces an alternating current that changes direction periodically, usually at a fixed frequency. The plot of its output voltage versus time would be a sinusoidal waveform with a repeating pattern of positive and negative values.
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. 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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Discuss all the components that make up the group of electromagnetic and electrical control equipment. Make sure you include the
following items: control relays, contactors, and motor starters. State the commonalities and major differences between these three
categories of control mechanisms. Expand your discussion with the implementation and adaptation of solid-state devices and how they
have changed the industrial and commercial electrical fields. Then make sure you integrate your discussion with comments about NEMA
and IEC-rated components.
Electromagnetic and electrical control equipment is used in a wide range of industrial and commercial applications to control the operation of electric motors, lighting, and other electrical loads.
What are the components of electromagnetic control equipment?Control relays, contactors, and motor starters are key components of electromagnetic and electrical control equipment.
They are used to control the operation of electric motors, lighting, and other electrical loads.
Solid-state devices have become increasingly popular in recent years due to their compact size, faster response times, and lower power consumption. NEMA and IEC establish standards for electrical equipment, and electrical components are often rated according to these standards.
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Please see Attached.
The total initial mechanical energy is 590,650 J.
Work done on the projectile is 155,436 J.
Speed of the projectile is 115.4 m/s.
How to calculate mechanical energy, work done and speed?(a) The initial total mechanical energy of the system of the projectile and the earth can be found using the equation:
E = KE + PE
where E is the total mechanical energy, KE is the kinetic energy, and PE is the potential energy. Initially, the projectile is at a height of 116 m, so the potential energy is:
PE = mgh
PE = (54.0 kg)(9.81 m/s²)(116 m)
PE = 61,450 J
The initial speed of the projectile is 1.40 x 10² m/s, so the initial kinetic energy is:
KE = (1/2)mv²
KE = (1/2)(54.0 kg)(1.40 x 10² m/s)²
KE = 529,200 J
Therefore, the initial total mechanical energy is:
E = KE + PE
E = 529,200 J + 61,450 J
E = 590,650 J
The initial total mechanical energy of the system of the projectile and the earth is 63,014 J.
(b) At the maximum height, the potential energy of the projectile is:
PE = mgy
PE = (54.0 kg)(9.81 m/s²)(320 m)
PE = 169,517 J
At the maximum height, the kinetic energy of the projectile is:
KE = (1/2)mv²
KE = (1/2)(54.0 kg)(99.2 m/s)²
KE = 265,697 J
The total mechanical energy of the projectile at the maximum height is:
E = KE + PE
E = 169,517 J + 265,697 J
E = 435,214 J
The work done by air resistance is the difference between the initial total mechanical energy and the total mechanical energy at the maximum height:
W = Ei - Ef
W = 590,650 J - 435,214 J
W = 155,436 J
The amount of work done on the projectile by air resistance is 155,436 J.
(c) Let Wup be the work done by air resistance when the projectile is going up, and let Wdown be the work done by air resistance when the projectile is going down. According to the problem, Wdown = (3/2)Wup. Use the principle of conservation of energy to relate the speed of the projectile at its maximum height to its speed just before hitting the ground:
Ef = Ei
KEf + PEf = KEi + PEi
At the maximum height, the kinetic energy of the projectile is zero, so:
PEf = Ei - PEi
PEf = 590,650 J - (54.0 kg)(9.81 m/s²)(320 m)
PEf = 421,133 J
The total mechanical energy just before hitting the ground is the same as the total mechanical energy at the maximum height, so:
KEf + PEf = KEi + PEi
0 + 421,133 J = (1/2)mv² + 61,450 J
(1/2)(54.0 kg)v² = 359,683 J
v² = 13,321.59
v = 115.4 m/s
Therefore, the speed of the projectile immediately before it hits the ground is 115.4 m/s.
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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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Calculate the mass of a solid iron sphere that has a diameter of 6.6 cm if its density is 7850 kg/m3
Answer:0.216 kg
Explanation:
The volume of a sphere is given by the formula:
V = (4/3)πr^3
where r is the radius of the sphere. Since the diameter of the iron sphere is 6.6 cm, its radius is half of that or 3.3 cm (0.033 m).
The volume of the sphere is:
V = (4/3)π(0.033 m)^3
V = 2.76 x 10^-5 m^3
The density of iron is 7850 kg/m^3. We can use the density formula to find the mass of the sphere:
density = mass / volume
mass = density x volume
mass = 7850 kg/m^3 x 2.76 x 10^-5 m^3
mass = 0.216 kg
Therefore, the mass of the solid iron sphere is 0.216 kg.
Answer:
Density of iron = 7870 kg/m3 diameter of sphere = 3.00 cm radius of sphere r = 1.5 cm =0.015 m Volume of sphere V = (4/3) * pi* (r^3) = (4/3) x
Explanation:
water of mass 120g at 50 degrees celcius should be added to 200g of water at 10 degrees celcius and the mixture is well stirred .calculate the temperature of the mixture[neglect heat losses to surrounding]
Answer:25 c
Explanation:
If a wave of wavelength 1m and another wave of wavelength 3m are both transmitted at a velocity of 120m/s, what is the frequency difference between these signals?
A 1300 kg car is driving east on 290 at 65 mph. What is the car's momentum?
The momentum of the car is 37,778.88 kgm/s in the east direction.
What is the car's momentum?Momentum is simply the product of the mass of an object and its velocity.
It is a vector quantity, which means that it has both magnitude (size) and direction.
Momentum = Mass × Velocity
in order to calculate the momentum of the car, we need to convert the velocity to meters per second (m/s) and also take into account the direction.
First, let's convert the velocity from mph to m/s.
We can do this by multiplying by a conversion factor:
65 mph × 0.44704 m/s per mph = 29.0576 m/s
So the velocity of the car is 29.0576 m/s in the east direction.
Now we can calculate the momentum of the car using the formula:
Momentum = Mass × Velocity
Momentum = 1300 kg * 29.0576 m/s east
Momentum = 37,778.88 kg m/s east
Therefore, the momentum of the car is 37,778.88 kgm/s.
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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.
From a point 120 m above ground level, a stone is projected vertically upwards with a speed of u m s. If the stone rises to a height of 5 m above the point of projection calculate (a) the value of u, (b) the time the stone takes from projection until it reaches ground level, (c) the speed of the stone at ground level.
If the stone rises to a height of 5 m above the point of projection, then the value of u is 49 m/s, the time the stone takes from projection until it reaches ground level is 5 seconds, and the speed of the stone at ground level is 98 m/s.
What is the calculation for speed?To find the value of u,
v = u + at (velocity (v) ,the initial velocity (u), acceleration (a), and time (t))
At the highest point, the velocity of the stone becomes zero,
0 = u - gt
t = u/g
this value of t is put into the equation for displacement,
5 m = 120 m + ut - [tex]0.5gt^2[/tex]
Substituting t = u/g,
5 m = 120 m + u(u/g) - [tex]0.5g(u/g)^2[/tex]
u = √(2gh) = √(2 x 9.8 [tex]m/s^2[/tex] x 115 m) = 49 m/s
b) v = u + at
-u = 49 m/s - gt
t = (49 m/s)/g + √(([tex]49 m/s)^2[/tex] + 4gh)/g
t = 5 s
c)
v = u + at
-u = 49 m/s - g(5 s)
u = 49 m/s + 9.8[tex]m/s^2[/tex] x 5 s
u = 98 m/s
Hence, the value of u is 49 m/s, the time the stone takes from projection until it reaches ground level is 5 seconds, and the speed of the stone at ground level is 98 m/s.
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Pleas Help
Questions are in the picture
1) There are many handwriting characteristics that are used to identify forgeries, including line quality, pen pressure, slope, letter size and shape, and more.
What is pressure ?Pressure is a physical force that is exerted on an object or area by another object or area. It is measured in terms of force per unit area and can be calculated using the formula: pressure = force/area. Pressure is an important factor in many areas of science, engineering, and everyday life.
2) Checks have embedded features such as watermarks, microprinting, and holograms to prevent forgery.
3) True. A person's handwriting is compared to several exemplars to determine if the handwriting is genuine or a forgery.
4) True. Some forgers use chemicals to age paper to make it appear older than it is.
5) The U.S. Department of the Treasury is charged with the security of U.S. currency.
6) False. People often spend time forging documents from famous people in order to make money.
7) U.S. currency has many security features, including watermarks, color-shifting ink, invisible fibers, and microprinting.
8) The main difference between check forgery and literary forgery is that check forgery is primarily concerned with creating or altering a financial document, while literary forgery is primarily concerned with creating or altering a work of literature.
9) Criminals make forged literary pieces look older by using aging techniques such as adding age spots
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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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how do you measure the amount of energy in each state of water
Water exists in three different states, namely solid ice, liquid and gaseous vapour. There are two different formulae, depending upon whether there is a loss or gain of heat, or change of state.
What is specific heat capacity and latent heat?The specific heat capacity is the amount of heat absorbed per unit mass of the substance for a temperature increase of one unit. Its units are given by J/kg K or cal/kg °C. It is useful in calculating the amount of heat lost or gained by a body (Q). The formula for Q is:
Q = m*c*ΔT
where, m is the mass
c is the specific heat capacity and,
ΔT is the change in the temperature of the body
This formula is used to measure the amount of energy required to raise or decrease the temperature of water without a change in its state.
On the other hand, latent heat is the energy released or absorbed when there is a change of state of matter. It is denoted by L. For a change of state, the heat energy Q is given by:
Q = m*L
Note that there is no change in temperature during change of state of matter, and only heat is released or absorbed.
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Please help (50 points and Brainly)
Answer:
0.0312J
Explanation:
Given:
Relaxed length of the spring (L0) = 0.115 mSpring constant (k) = 51.0 N/mExtended length of the spring (L) = 0.150 mWant: Elastic potential energy stored in the spring
Solve:
The elastic potential energy stored in the spring can be calculated using the following formula:
Elastic potential energy = 0.5 * k * (L - L0)^2
Substituting the given values into the equation:
Elastic potential energy = 0.5 * 51.0 N/m * (0.150 m - 0.115 m)^2
Elastic potential energy = 0.5 * 51.0 N/m * (0.035 m)^2
Elastic potential energy = 0.5 * 51.0 N/m * 0.001225 m^2
Elastic potential energy = 0.0312 J
Therefore, the elastic potential energy stored in the spring when its length is 0.150 m is 0.0312 J.
A toy cannon with a mass of 1.0 kg is initially at rest on a horizontal, frictionless surface. It shoots a 0.05 kg ball with a velocity of 10 cm/s at an angle of 30° above the horizontal. What is the speed of the 1.0 kg cannon immediately after the projectile is released?
Here, the total momentum before and after the release of the projectile is same. From this concept the final velocity of the cannon after the release will be 0.00433 m/s.
What is conservation of momentum ?Momentum of an object is the product of its mass and velocity. During a collision, the momentum after and before the collision is constant. This can be applied in the case of a projectile from the cannon.
Here, the initial momentum of the cannon and the ball is zero, because they are at rest.
After, projectile the total momentum momentum is zero itself according to conservation of momentum.
momentum of the ball = 0.05 kg × 0.1 m/s cos 30 + 1 kg × v = 0
Here v is the velocity of the cannon.
then v = 0.05 kg × 0.1 m/s cos 30 / 1 kg
= 0.0043 m/s.
Therefore, the velocity of the cannon after the projectile is released is 0.0043 m/s.
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a boy pushes a lever down 2 meters with a force of 75 newtons. The box at the other end with a weight of 50 newtons move up 2.5 meters
A boy pushes a lever down 2 meters with a force of 75 newtons. The box at the other end with a weight of 50 newtons move up 2.5 meters. The efficiency of this machine is 83.33%.
What is efficiency?The energy lost during a machine's operation due to heat and friction is measured as the machine's efficiency. Given that work is the transformation of kinetic energy, a machine's efficiency can be expressed as the ratio of output work divided by input work less work wasted due to friction and heat. It is unitless.
Using the formula:
Efficiency=[tex]\frac{O}{I}[/tex]×100
where,
O is output work and,
I is the input work.
Work done= Force × displacement
Output work = 50 × 2.5 = 125J
Input work = 75 × 2 = 150J
Substituting the values and solving for efficiency:
Efficiency = 83.33%
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Complete question:
A boy pushes a lever down 2 meters with a force of 75 Newtons. The box on the other end weighs 50 Newtons and moves up 2.5 meters. What is the efficiency of this machine?
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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A current of 4.5 A flows through a point for 25 minutes. Calculate the charge through the point after 25 minutes.
Answer:
I=4.5A,t=25minutes Q=?
We know that I=Q/T
Q=25×4.15=
112.5c
Explanation:
which of the following is an example of uniform acceleration
motion under force
motion under gravity
motion under power
none
The following is example of uniform acceleration, that is : motion under force.
What is uniform acceleration?Uniform acceleration refers to constant rate of change in velocity, which means that acceleration of an object is constant over time. Motion under force can result in uniform acceleration if the force is constant and object is free to move in straight line.
Motion under gravity and motion under power can both result in acceleration but they are not necessarily uniform. For example, object falling under gravity experiences constantly increasing acceleration due to increasing speed and gravitational force acting on it, which means acceleration is not uniform.
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A particle is projected vertically upwards from the ground with a speed of 36m/s^-1, calculate.
a. The time for which it is above a height of 63m.
b. The speed which it has at this height on its way down
c. The total time of flight
To solve this problem, we can use the equations of motion for a freely falling body with constant acceleration due to gravity (g = 9.81 m/s²).
a. The time for which it is above a height of 63m:
We can use the following equation of motion to find the time t it takes for the particle to reach a height of 63 m above the ground:
y = uyt + (1/2)gt²
where y is the displacement (height) of the particle, uy is the initial velocity in the y-direction (upward), and t is the time.
At the highest point of its trajectory, the particle will have zero velocity, so we can use the initial velocity as uy = 36 m/s. We want to find the time t when the particle reaches a height of y = 63 m, so we can rearrange the equation to solve for t:
t = (-uy ± sqrt(uy² + 2gy))/g
where the ± sign indicates that there are two possible solutions, one for when the particle is going up (positive root) and one for when the particle is coming down (negative root).
Plugging in the values for uy, g, and y, we get:
t = (-36 ± sqrt(36² + 2*(-9.81)*63))/(-9.81)
t = 6.53 s (upward)
b. The speed which it has at this height on its way down:
When the particle reaches a height of 63 m on its way down, we can use the following equation to find its velocity vy:
vy² = uy²+ 2g(y - yo)
where yo is the initial height (0 m) and uy is the initial velocity in the y-direction (upward).
We can plug in the values for uy, g, and y, and solve for vy:
vy²= 36²+ 2(-9.81)(63 - 0)
vy = 32.2 m/s (downward)
c. The total time of flight:
The total time of flight is twice the time it takes for the particle to reach the maximum height, since it takes the same amount of time to come back down to the ground. Therefore, the total time of flight is:
t_total = 2t_upward = 2(6.53) = 13.06 s
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URGENT HELP NEEDED!!! BRAINLIEST WILL BE PICKED!!!
Two magnets are sliding towards each other on a smooth surface. Magnet A, which has a mass of 288 g, is moving in the negative x-direction at a speed of 4.8 m/s. Magnet B, which has a mass of 112 g, is moving in the positive x-direction at a speed of 7.8 m/s. When they collide, they join together, and move as a single unit. What is the velocity of this single unit after the collision?
−1.3 m/s
5.6 m/s
1.3 m/s
−5.6 m/s
Answer:
1.3 m/s
Explanation:
You throw a water ballon straight up with a velocity of 12 m/s. What is its maximum height
Answer:7.35 meters
Explanation:
Assuming that air resistance is negligible, we can use the kinematic equation:
h = (v^2)/(2g)
where h is the maximum height, v is the initial velocity, and g is the acceleration due to gravity (9.8 m/s^2).
Substituting the given values:
h = (12^2)/(2 × 9.8)
h = 7.35 meters
Therefore, the maximum height of the water balloon is approximately 7.35 meters.
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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URGENT HELP NEEDED!!!
A gardener uses a shovel as a lever to lift a 200 N rock a distance of 0.20 meters. He does this by applying 50 N of force to the end of the shovel. a) Calculate the mechanical advantage of the lever. b) How far down does the gardener push the shovel?
a) 4 b) 0.8 m
a) 4 b) 0.4 m
a) 40 b) 1.2 m
a) 2 b) 0.8 m
The mechanical advantage of the lever will be 4.
The distance the gardener pushed the shovel will be 0.8 m.
Work done problema) The mechanical advantage of the lever can be calculated using the formula:
MA = output force/input force
In this case, the output force is the weight of the rock (200 N) and the input force is the force applied to the shovel (50 N). Therefore:
MA = 200 N / 50 N = 4
So the mechanical advantage of the lever is 4.
b) The distance that the gardener pushes the shovel can be calculated using the formula:
output distance / input distance = input force / output force
In this case, the output distance is the distance that the rock is lifted (0.20 m) and the input distance is the distance that the gardener pushes the shovel (unknown). We know the input force (50 N) and the output force (200 N), so we can set up the equation:
0.20 m / x = 50 N / 200 N
Simplifying and solving for x:
0.20 m * 200 N / 50 N = x
0.8 m = x
Therefore, the gardener pushes the shovel down 0.8 m.
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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
HELP PLEASEEEEEE HELP PLEASE
The letter that represents the various descriptions is as follows:
5.) Angle of incidence: B
6.) Object: D
7.) Plain mirror:A
8.) Reflection: E
9.) Angle of reflection: C
10.) Normal: F
How does the diagram explain the law of reflection?The law of reflection in a plain mirror states that the angle of reflection is equal to the angle of incidence.
The diagram explains the law of reflection on a plain mirror because it shows the reflection of the object D as E after creating an angle of incidence B on the plane mirror surface A which is equal to the angle of reflection C.
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