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Morgarella [4.7K]
3 years ago
14

HOW MANY PIES DOSE IT TAKE TO GET TO THE MOON

Physics
1 answer:
Schach [20]3 years ago
8 0

i do not have an answer because it depends on the size and the distance lol

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A 10.0 kilogram mass of iron on earth exerts a downward force of 98 Newtons. On the moon this same 10.0 kilograms of iron is wei
Len [333]

Answer: D) The gravity on the moon is one sixth the gravity of Earth.  

Explanation:

6 0
3 years ago
Read 2 more answers
50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t
r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

KE=\frac{1}{2}mv^2

where v is the speed of the object

When the ball is sitting on the top of the building, we have

  • h=40 m, therefore the potential energy is not zero
  • v=0, since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

h=20 m

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

E=PE+KE=const.

At the top of the building,

E=PE_{top}

While halfway through the fall,

PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}

And the mechanical energy is

E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}

which means

KE_{half}=\frac{E}{2}

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

  • The height of the ball relative to the ground is now zero: h=0. This means that the potential energy of the ball is zero: PE=0
  • The kinetic  energy, instead, is not zero: in fact, the ball has gained speed during the fall, so v\neq 0, and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

PE=(2)(9.8)(40)=784 J

5)

The potential energy of the ball as it is halfway through the fall is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

PE=(2)(9.8)(20)=392 J

6)

The kinetic energy of the ball halfway through the fall is given by

KE=\frac{1}{2}mv^2

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the  fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

KE=\frac{1}{2}(2)(19.8)^2=392 J

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

KE=\frac{1}{2}mv^2

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

KE=\frac{1}{2}(2)(28)^2=784 J

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0
3 years ago
HOW DO U FEEL WHEN U PLAY OR WATCH BADMINTON?
True [87]

Answer:

I feel exited and happy I enjoy it with my friend

6 0
3 years ago
In a collision, a 25.0 kg mass moving at 3.0 m/s transfers all of its momentum to a 5.0 kg mass.
nadezda [96]

Answer:

Explanation:

The momentum of the 25 kg mass is

p=mv

p=25kg*3m/s= 75kg*m/s

If this whole momentum of the object is transferred to the 5.0 kg object then according to the law of conservation of momentum, the momentum of the 25.0 kg object must be transferred to the 5.0 kg object:

75kg*m/s = 5.0kg*v

v=\dfrac{75}{5}

\boxed{v=15m/s}

8 0
3 years ago
Transform boundaries are classified under which type of fault?
cluponka [151]

Answer:

Strike-slip fault

Explanation:

Transform boundaries play the role of connecting the other plate boundary segments.

When the plates are rubbed against each other, they result in enormous amount of stresses which leads to the breaking of the part of a rock causing earthquakes. Places of occurrence of these breaks are termed as faults.

Strike slip faults results from compression which takes place horizontally, but but in this the rock displacement  releases energy and takes place in a horizontal direction which is parallel to the force of compression.

8 0
3 years ago
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