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mash [69]
3 years ago
5

What is wrong with the following statement: When you exert a force on a baseball, the equal and opposite force on the ball balan

ces the original force and therefore, the ball will not accelerate in any direction.
Physics
1 answer:
gregori [183]3 years ago
6 0

Answer:

When you exert a force on a baseball, there exists an equal and opposite force on the ball therefore, the ball will accelerate in opposite direction.

Explanation:

When you hit a ball with baseball bat, the bat exerts a great force on the ball which causes the ball to accelerate in the opposite direction. It is to be noted that the mass of bat is much greater than mass of ball but the acceleration of ball is also greater than the acceleration of the bat so both bat and ball almost exert same magnitude of force but in opposite direction and as a result both bat and ball accelerate in opposite direction, the deciding factor is of course the relative force applied by the batter and the bowler.

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How far an object moves from a reference point is known as<br> type your answer...
fenix001 [56]

Answer:

distance

hope it helps

Explanation:

-

5 0
3 years ago
How much heat is needed to raise the temperature of 5g of water by 20oC?
Ratling [72]
Required Heat = Q

Q = Mass * specific heat of water * change in temp.

Q = 5g * 1g/cal*degC * 20degC

Q = 100 cal of heat is required

To convert calories to Joules,
1 cal = 4.184 Joules

100cal = 418.4 J of heat is needed
5 0
3 years ago
A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
Angelina_Jolie [31]

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

a)

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

\alpha=\frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time elapsed

The angular velocity can be written as

\omega=\frac{2\pi}{T}

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

T'=T+8.09\cdot 10^{-6} =0.01400809 s

Therefore, the change in angular velocity after 1 year is

\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s

So, the angular acceleration of the pulsar is

\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y

b)

To solve this part, we can use the following equation of motion:

\omega'=\omega + \alpha t

where

\omega' is the final angular velocity

\omega is the initial angular velocity

\alpha is the angular acceleration

t is the time

For the pulsar in this problem:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s is the initial angular velocity

\omega'=0, since we want to find the time t after which the pulsar stops rotating

\alpha = -0.259 rad/s/y is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y

c)

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y

which means that the period of the pulsar increases by

\Delta T=8.09\cdot 10^{-6} s

For every year:

\Delta t=1 y

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

T=T_0+\frac{\Delta T}{\Delta t}\Delta t

where T_0 is the original period and

\Delta t=869 y

is the time that has passed; solving for T0,

T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s

6 0
3 years ago
Please Help
satela [25.4K]
A man pulls on a box with 50 N of applied force while another man pushes on the box from the opposite direction with 60 N of applied force. The box will stand still.

TRUE.
4 0
3 years ago
Read 2 more answers
A certain string just breaks when it is under400N of tension.
OLEGan [10]

Answer:

The string will break with a speed of 20 m/s.

Explanation:

It is given that,

Tension at which the string just breaks, T = 400 N

Mass of the stone, m = 10 kg

Radius of the circle, r = 10 m

We need to find the speed at which the string will break. The boy continuously increases the speed of the  stone. The tension acting on the stone is equal to the centripetal force. It is a force that acts towards the center of circle. It is given by :

T=F=\dfrac{mv^2}{r}

v=\sqrt{\dfrac{Tr}{m}}

v=\sqrt{\dfrac{400\ N\times 10\ m}{10\ kg}}

v = 20 m/s

So, the string will break with a speed of 20 m/s. Hence, this is the required solution.

4 0
3 years ago
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