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Anton [14]
3 years ago
15

Directions: For each situation, you need to determine if the object has KE or GPE or both. Then find their values. Be sure to in

clude the correct units.
A 2-kg bowling ball sits on top of a building that is 40 meters tall.



Circle one: KE / GPE / both



Show your work for finding the values of each type of energy the object has:









A 2-kg bowling ball rolls at a speed of 5 m/s on the roof of the building that is 40 tall.



Circle one: KE / GPE / both



Show your work for finding the values of each type of energy the object has:







A 2-kg bowling ball rolls at a speed of 10 m/s on the ground.



Circle one: KE / GPE / both



Show your work for finding the values of each type of energy the object has:







A 20-kg child is tossed up into the air by her parent. The child is 2 meters off the ground




traveling 5 m/s.



Circle one: KE / GPE / both



Show your work for finding the values of each type of energy the object has:



























Part 2: Finding Speed and Height Using Energy



Directions: Use the KE and GPE equations to solve for speed or height depending on the situation.



A 1,000-kg car has 50,000 joules of kinetic energy. What is its speed?





A 200-kg boulder has 39,200 joules of gravitational potential energy. What height is it at?





A 1-kg model airplane has 12.5 joules of kinetic energy and 98 joules of gravitational potential energy. What is its speed? What is its height?
Physics
2 answers:
FromTheMoon [43]3 years ago
5 0


 A 2-kg bowling ball sits on top of a building that is 40 meters tall.

GPE

A 2-kg bowling ball rolls at a speed of 5 m/s on the roof of the building that is 40 tall

its moving n its on top of a bldg. so ans is both KE n GPE

A 2-kg bowling ball rolls at a speed of 10 m/s on the ground.

KE

A 20-kg child is tossed up into the air by her parent. The child is 2 meters off the ground traveling 5 m/s.

moving n in the air. ans is both KE n GPE


eqn:  KE= 1/2mv^2, GPE=hmg

A 1,000-kg car has 50,000 joules of kinetic energy. What is its speed?

50000=1/2*1000^v^2

v=10m/s


 A 200-kg boulder has 39,200 joules of gravitational potential energy. What height is it at?

39200=200*9.8*h

h=20m


A 1-kg model airplane has 12.5 joules of kinetic energy and 98 joules of gravitational potential energy. What is its speed? What is its height?

12.5=1/2*1*v^2

v=5m/s

98=1*9.8*h

h=10m


Liula [17]3 years ago
5 0
Part 1

First  GPE
Second  both KE n GPE
Third  KE
Fourth  both KE n GPE


A 1,000-kg car has 50,000 joules of kinetic energy.

1/2*1000^speed^2=50000
speed=10m/s

 A 200-kg boulder has 39,200 joules of gravitational potential energy.
200*9.8*height=39200
height=20m

A 1-kg model airplane has 12.5 joules of kinetic energy and 98 joules of gravitational potential energy.

1/2*1*speed^2=12.5
speed=5m/s
1*9.8*height=98
height=10m
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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

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3 years ago
A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a
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Answer:

4 hoop, disk, sphere

Explanation:

Because

We are given data that

Hoop, disk, sphere have Same mass and radius

So let

And Initial angular velocity, = 0

The Force on each be F

And Time = t

Also let

Radius of each = r

So let's find the inertia shall we!!

I1 = m r² /2

= 0.5 mr² the his is for dis

I2 = m r² for hoop

And

Moment of inertia of sphere wiil be

I3 = (2/5) mr²

= 0.4 mr²

So

ωf = ωi + α t

= 0 + ( τ / I ) t

= ( F r / I ) t

So we can see that

ωf is inversely proportional to moment of inertia.

And so we take the

Order of I ( least to greatest ) :

I3 (sphere) , I1 (disk) , I2 (hoop) , ,

Order of ωf: ( least to greatest)

That of omega xf is the reverse of inertial so

hoop, disk, sphere

Option - 4

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Answer:

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Explanation:

Given data,

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The distance the ball rolled, d = 24 m

The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,

                              v = d / t      m/s

Substituting the given values in the above equation,

                               v = 24 / 8

                                  = 3 m/s

Hence, the speed of the ball was, v = 3 m/s

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