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A 21 kg mountain lion carries a 3kg cub in it's mouth as it jumps from rest on the ground to the top of a 2 m talk rock. It takes 1 seconds for the mountain lion to jump and reach the top. How much power did the mountain lion exert? I need help to solve for power

Answers

Answer 1
Answer:

Answer:

The power exerted by the mountain lion is 1,472.35 W.

Explanation:

Given;

mass of mountain, m₁ = 21 kg

mass of the cub, m₂ = 3 kg

height jumped by the mountain lion, h = 2 m

time taken for the mountain lion to jump, t = 1 s

Determine the weight of the lions on the top rock;

W = F = (m₁ + m₂)g

F = (21 + 3) x 9.8

F = (24) x 9.8

F = 235.2 N

Determine the final velocity of the mountain rock as it jumped to the top;

v² = u² + 2gh

where;

u is the initial velocity = 0

h is the height jumped = 2 m

v² = 0 + 2 x 9.8 x 2

v² = 39.2

v = √39.2

v = 6.26 m/s

The power exerted by the mountain lion is calculated as;

P = Fv

P = 235.2 x 6.26

P = 1,472.35 W

Therefore, the power exerted by the mountain lion is 1,472.35 W.

Related Questions

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. If g-9.8 m/s, what is the coefficient of static friction? a. 3.0 b. 0.15 c. 0.28 d. 0.31
An earthquake 45 km from a city produces P and S waves that travel outward at 5000 m/s and 3000 m/s, respectively. Once city residents feel the shaking of the P wave, how much time do they have before the S wave arrives in seconds?
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Chuck wants to investigate how gas moleculesmove in a container. Which model would be most
helpful to represent this motion?
A. stacking blocks to build a tower
B. freezing water in an ice cube tray
C. bouncing elastic balls off of each other and
the walls of a room
D. placing a closed, water-filled plastic bag in
the sun and watching condensation form

Answers

The answers C the molecules in gas move rapidly and all around they are spread out and bounce off each-other

A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the lowest frequency associated with this portion of the spectrum?

Answers

Answer:

the lowest frequency is 7.5* 10^(14) Hz

Explanation:

In the question it is given that wavelength(L) in the range of 200μm to 400μm.

let ν be frequency of wave v velocity = 3\times 10^8

velocity v= Lν

therefore ν=(v)/(L)

frequency ν be lopwest when L will be heighest

ν(lowest)=(3* 10^8)/(400* 10^-9)

ν=7.5* 10^(14) Hz

In straight line motion, if the velocity of an object is changing at a constant rate, then its position is _________ and its acceleration is___________: O changing: zero O changing; changing O constant and non-zero; constant and non-zero O None of the above

Answers

Answer:

None of the above

It should be position is changing and acceleration is constant.

Explanation:

Since the velocity is changing, this means the object is moving, so the position must also be changing.

Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the acceleration must be constant too.

So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.

So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"

Final answer:

In straight line motion, if velocity changes at a constant rate, then the position is changing and the acceleration is constant and non-zero. This is defined under the principles of kinematics and implies that as the velocity alters constantly, the object is in motion, hence its position is changing.

Explanation:

In straight line motion, if the velocity of an object is changing at a constant rate, then its position is changing and its acceleration is constant and non-zero. This condition is defined under the laws of physics, more specifically, under the study of kinematics.

The acceleration is constant because you're considering a situation where velocity is changing at a constant rate. In this case, the change in velocity is the acceleration, which is a constant and not zero. This situation is described by the kinematic equations for constant acceleration.

The position is changing because the object is moving. A change in position over time constitutes motion, and in this case, because the velocity (the rate of change of position) is changing, the object's position cannot be constant.

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In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.

Answers

Answer:

Explanation:

Let length of the pendulum be l . The expression for time period of pendulum is as follows

T = 2π\sqrt{(l)/(g) }

For Mars planet ,

1.5 = 2\pi\sqrt{(l)/(.38*9.8) }

For other planet

.92 = 2\pi\sqrt{(l)/(g_1) }

Squiring and dividing the two equations

(1.5^2)/(.92^2) = (g_1)/(3.8*9.8)

g_1 = 9.9

The second planet appears to be earth.

A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the
ball?

Answers

Answer:

v = 0.85 m/s

Explanation:

Given that,

Mass of the ball, m = 0.01 kg

Centripetal force on the ball, F = 0.025 N

Radius of the circular path, r = 0.29 m

Let v is the speed of the ball. The centripetal force of the ball is given by :

F=(mv^2)/(r)

v=\sqrt{(Fr)/(m)}

v=\sqrt{(0.025* 0.29)/(0.01)}

v = 0.85 m/s

So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.

An internal explosion breaks an object, initially at rest,intotwo pieces, one of which has 1.5 times the mass of the other.If
7500 J were released in the explosion, how much kinetic energydid
each piece acquire?

Answers

Answer:

4500 J and 3000 J

Explanation:

According to conservation of momentum

      0 = m_1 V_1 + m_2 V_2

Given that m_2 = 1.5 m_1 , so

    V_1 = -1.5 V_2

  the kinetic energy of each piece is

    K_2= (1)/(2) m_2v_2^2

    K_1= (1)/(2) m_1v_1^2

substituting the value of V1 in the above equation

    K_1 = (1/2)( m_2 / 1.5 )( -1.5 V_2)^2 = 1.5 (1/2)m_2 V_2^2 = 1.5 K_2

  Given that

         K_1 + k_2 = 7500 J

       1.5 K_2 + K_2 = 7500

         K_2 = 7500 / 2.5

               = 3000 J

this is the KE of heavier mass

      K_1 = 7500 - 3000 = 4500 J

this is the KE of lighter mass

Final answer:

The question is about finding the kinetic energy acquired by each of two pieces of an object following an internal explosion, using principles of conservation of energy and momentum in physics.

Explanation:

The student has asked about an internal explosion that breaks an object into two pieces with different masses, releasing a certain amount of kinetic energy in the process. This question involves applying the principle of conservation of energy and momentum to find the kinetic energy acquired by each piece post-explosion.

Assuming piece 1 has a mass of m and piece 2 has a mass of 1.5m, the total mass of the system is 2.5m. Since 7500 J of energy was released in the explosion, to find the kinetic energy of each piece, we can use the fact that the total kinetic energy is equal to the energy released during the explosion. Let the kinetic energy of the smaller piece be K1 and of the larger piece be K2. Because the object was initially at rest and momentum must be conserved, the momenta of the two pieces must be equal and opposite. This relationship allows us to derive the ratio of the kinetic energies. We can solve for K1 and K2 proportionally. Finally, because the kinetic energy is a scalar quantity, adding the kinetic energies of the two pieces will equal the total energy released.

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