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4 years ago

A batter hits a baseball with a bat. The bat exerts a force on the ball. Does the ball exert a force on the bat?

Physics

2 answers:

pashok25 [27]4 years ago

8 0

Answer:

when batter hit the ball with bat then force is applied by the bat on the ball and same force is applied on the bat by the ball.

This is an example of action reaction law

Explanation:

As we know by newton's III law that every action has equal and opposite reaction force

Which means if a force is applied by an object to another object then the same magnitude force is applied by that object on the first one.

The force will be of same magnitude but opposite in direction

This is known as reaction force.

As per Newton's III law every action has equal and opposite reaction force on it.

So here when batter hit the ball with bat then force is applied by the bat on the ball and same force is applied on the bat by the ball.

This is an example of action reaction law

Mazyrski [523]4 years ago

3 0

Yes, think about the difference of swinging a bat and not hitting a ball. It's fairly easy right? Now, when you hit a ball with the bat, you will feel the bat sting your hands. That's the force the ball is exerting on the bat!

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A horizontal spring required a force of 1.0 N to compress it 0.1 m. How much work is required to stretch the spring 0.4 m?

Crank

Answer:

Explanation:

given data

Force applied F= 1N

extension e= 0.1m

let us find the spring constant first

applying

F=ke

k=F/e

k=1/0.1

k=10N/m

Step two:

Required is the work done

we know that the expression/formula for the work done by a spring is given as

Wd=1/2kx^2

x=0.4m

substitute

Wd= 1/2*10*0.4^2

Wd=0.5*10*0.16

Wd=0.5J

3 0

3 years ago

If we decrease the distance an object moves we will

Scrat [10]

Decrease the amount of work done.

4 0

3 years ago

Read 2 more answers

The latent heat of vaporization for water at room temperature is 2430 J/g.

alukav5142 [94]

Answer:

1) $kinectic energy=7.26*10^-^2^0J$

2) $V= 2.0m/s$

3) $T=3.5*10^3K$

4) The Molecules do not burn because of the presences of hydrogen bond in place

Explanation:

From the question we are told that

latent heat of vaporization for water at room temperature is 2430 J/g.

1)Generally in determining the molar mass of water evaporated we have that

-One mole (6.02 x 10. 23 molecules)

-Molar mass of water is 18.02 g/mol

Mathematically the mass of water is give as

$M=\frac{18.02}{6.02*10^-^2^6}$

$M=3*10^-^2^3g$

Therefore

$kinectic energy=2430J/g*3*10^-^2^3g$

$kinectic energy=7.26*10^-^2^0J$

b)Generally the evaporation speed V is given as $V= \sqrt{\frac{K.E*2}{m} }$

Mathematically derived from the equation

$\frac{1}{2} mv^2 =K.E$

To Give

$V= \sqrt{\frac{K.E*2}{m} }$

$V= \sqrt{\frac{7.26*10^-^2^0J*2}{3*10^-^2^3g} }$

$V= 2.0m/s$

c)Generally the equation for velocity $Vrms=\sqrt{\frac{3RT}{M} }$

Therefore

Effective temperature T is given by

$T=\frac{\sqrt{v}*m}{R}$

where

$T=\frac{\sqrt{2.0m/s}*6.02*10^-^2^6}{0.082057 L atm mol-1K-1}$

$T=3.5*10^3K$

4) The Molecules do not burn because of the presences of hydrogen bond in place

3 0

3 years ago

The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

$|\Delta E_{ij}| = |E_i-E_j|$

Our states are given by

$E_1 = -11.7eV$

$E_2 = -4.2eV$

$E_3 = -3.3eV$

In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

$|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV$

$|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV$

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0

4 years ago

Difference between incident ray and refracted ray

natima [27]

Answer:

** incident ray.

Incident ray - the ray of light falling on the surface AB is called the incident ray

reflected ray.

** Reflected ray - the incident ray bouncing back in the same medium after striking the reflecting surface is called reflected ray.

3 0

4 years ago