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netineya [11]
3 years ago
10

PLEASEEEEEE HELLPPPP

Physics
1 answer:
malfutka [58]3 years ago
5 0

Answer: one of them is more lighter to move  and the other one is more heavy to move. hope this helps

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Which best describes the motion of air particles when a transverse wave passes through them?
Minchanka [31]
C.
The particles move perpendicular to the direction of the wave.
3 0
3 years ago
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 you calculate the mass of an object accelerating at22.35m/s2 with a force of 120N
Arte-miy333 [17]

Answer:

The mass of object is calculated as 5.36 kg

Explanation:

The known terms to find the mass are:

           acceleration of object (a) = 22.35 m/s^{2}

                        Force exerted (F) = 120N

                        mass of an object (m) = ?

From Newton's second law of motion;

                                   F = ma

                           or, 120 = m × 22.35

                          or, m= \frac{120}{22.35} kg

                           ∴ m = 5.36 kg

3 0
3 years ago
A vacuum cleaner has a rating of 460 W on 230 V mains. The value of the fuse connected in the plug will be
amid [387]

Answer:

2 amps

Explanation:

Given data

Power = 460W

voltage= 230V

Required

The amperage/ current of the fuse

Recall  P= IV

I= P/V

I= 460/230

I=2 amps

Hence the current of the fuse is 2 amps

7 0
3 years ago
your friend sit in a sked in the snow. if you apply a force of 120 N to them, they have an acceleration of 1.3 m/s2. what is the
levacccp [35]
Need to know the equation for force
F=MA
F is force
M is mass- we need to know the mass
A is acceleration
use "x" for mass
120 N= 1.3x
divide 1.3 in both side
kg unit for mass
X=92.31 kg 
or 
mass = 92.31 kg
Hope this helps
7 0
4 years ago
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