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Dima020 [189]
4 years ago
8

A small dog is trained to jump straight up a distance of 1.2 m. How much kinetic energy does the 7.2-kg dog need to jump this hi

gh? (The acceleration due to gravity is 9.8 m/s2.) Show your work for credit.
Physics
1 answer:
LUCKY_DIMON [66]4 years ago
8 0
Answer:84.672 joules.

Explanation:

1) Data:

m = 7.2 kg
h = 1.2 m
g = 9.8 m / s²

2) Physical principle

Using the law of mechanical energy conservation principle, you have that the kinetic energy of the dog, when it jumps, must be equal to the final gravitational potential energy.

3) Calculations:

The gravitational potential energy, PE, is equal to m × g × h

So, PE = m × g × h = 7.2 kg × 9.8 m/s² × 1.2 m = 84.672 joules.

And that is the kinetic energy that the dog needs.
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A 143 kg astronaut is floating in
Dima020 [189]

The recoil velocity of the astronaut is -0.070 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the astronaut-wrench system must be conserved.

At the beginning, their total momentum is zero:

p=0 (1)

Later, after the astronaut throws the wrench, the total momentum is

p=mv+MV (2)

where

m = 0.725 kg is the mass of the wrench

v = 13.8 m/s is the velocity of the wrench

M = 143 kg is the mass of the astronaut

V is the recoil velocity of the astronaut

Since momentum is conserved, (1) = (2), and so we can find V:

0=mv+MV\\V=-\frac{mv}{M}=-\frac{(0.725)(13.8)}{143}=-0.070 m/s

Learn more about momentum:

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brainly.com/question/9484203

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3 years ago
Imagine a large block that sinks in a tub of water. You hold a much smaller block of the exact same material in your hand, above
goldenfox [79]
It will sink because mass does not affect the physical properties of the object.
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What happens when a force is applied to an object in stable equilibrium
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Answer:

the object is no longer in equilibrium .

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List all the forces that act on a hot air ballon when it is in the sky
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A 320 g air track cart traveling at 1.25 m/s collides with a stationary 270 g cart. What is the speed of the 270 g cart after th
Nutka1998 [239]

Answer:

The speed of the 270g cart after the collision is 0.68m/s

Explanation:

Mass of air track cart (m1) = 320g

Initial velocity (u1) = 1.25m/s

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Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s

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3 years ago
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