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Len [333]
2 years ago
11

What average net force is required to accelerate a 3950 kg bus to a speed of 25m/s in 10.5 s?

Physics
1 answer:
tatuchka [14]2 years ago
6 0

Answer:

Explanation:

ACCELERATION = <u>CHANGE IN VELOCITY </u>

                                          TIME  TAKEN

∴ a = <u>25 - 0 </u>

         10.5

∴ a =  <u> 50  </u>  

           21    

by using f = ma,

we get , f =   3950 ×  <u>50 </u>                                                                                  

                                     21

∴  A FORCE OF 9404.8 N IS NEEDED TO ACCELERATE A BUS OF 3950 KG.

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What mass will accelerate at 3 m/s² when a net force of 150 N acts on it?
vodomira [7]

Answer:

\huge\boxed{\sf m = 50\ kg}

Explanation:

<h3><u>Given data:</u></h3>

Acceleration = a = 3 m/s²

Force = F = 150 N

<h3><u>Required:</u></h3>

Mass = m = ?

<h3><u>Formula:</u></h3>

F = ma

<h3><u>Solution:</u></h3>

Put the givens in the formula

150 = m (3)

Divide 3 to both sides

150/3 = m

50 kg = m

m = 50 kg

\rule[225]{225}{2}

7 0
1 year ago
The force exerted on the tires of a car that directly accelerate it along a road is exerted by the
azamat

The force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

<h3>What is force?</h3>

Force is defined as the product of mass and acceleration of an object.

Friction is defined as the force that resists the movement of an object over another.

Therefore, the force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

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7 0
2 years ago
What is the five things to do in P.E
Nimfa-mama [501]

Answer:

Fucd6

Explanation:

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3 years ago
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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.

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4 0
3 years ago
Forces always act in pairs? True or false
nasty-shy [4]

Answer:

True

Explanation:

Just as Isaac Newton says, "For every action, there is an equal and opposite reaction."

7 0
3 years ago
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