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Alinara [238K]
3 years ago
13

According to Newton’s Third Law of Motion, how are action and reaction forces related. Provide a full explanation with an exampl

e
Physics
1 answer:
hammer [34]3 years ago
5 0
For every action, there is an equal and opposite reaction. For example, take a cup on a table. The weight of the cup is the action, and the reason the cup does not sink through the table is because the table exerts an equal reaction force which is opposite to the action of the cup.

Hope this helps!
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Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.
Semmy [17]

Answer:

Explanation:

Given

time of contact between bat and ball is t=0.00129 s

suppose u is the incoming velocity and v is the final velocity after collision

m=mass\ of\ ball

Impulse exerted is given by change in momentum of the particle.

Initial momentum P_i=m\times u

Final momentum P_f=m\times v

Change in momentum \Delta P=P_i-P_f

Impulse    J=F_{avg}\cdot t=\Delta P

J=F_{avg}\cdot t=m(u-v)

F_{avg}=\frac{m(v-u)}{t}

F_{avg}=775.2\times m(v-u)\ N                

5 0
3 years ago
Cellular telephones use what type of wave?
lana [24]
Cellphones use radio frequency waves to transmit sound through the speaker and the microphone.<span> The radio frequency energy that is given off by cell phones is a type of electromagnetic energy.
I hope I helped! =D</span>
4 0
2 years ago
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The sun can continue to exist in its present stable state for about another
Wewaii [24]

Answer:

5.5 billion years

Explanation:

The answer

3 0
2 years ago
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assuming birdman flies at height of 72m, how fast should he fly to hit bucket at 63m from start of field. gravity is -9.8m/s^2 n
algol [13]

Answer:

.....Birdman?

Explanation:

7 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
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