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3 years ago

If the net force on an object is zero, what is the acceleration?

Physics

1 answer:

Finger [1]3 years ago

5 0

Answer:

The net force acting on the body is zero, the acceleration acting on the object is also zero.

Explanation:

Let the two equal and opposite forces acts on an object be, f and f'

The net force acting on the object is given by the relation,

The net force is given by the resultant vector,

$F= \sqrt{f^{2} + f'^{2} + 2 ff' Cos \theta}$

$F= \sqrt{f^{2} - f'^{2} + 2 ff'$ (∵ θ = 180)

$F=\sqrt{(f - f')^{2} }$

F = f - f'

Since the magnitude of the forces are equal,

F = 0

Hence, the net force acting on the body is zero, the acceleration acting on the object is also zero.

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3 years ago

All stored energy is potential<br>energy. Do you agree?<br>

umka21 [38]

Answer:

yes I agree Because........

Explanation:

Potential energy is stored energy. Potential energy is the energy that exists by virtue of the relative positions of the objects within a physical system. This form of energy has the potential to change the state of other objects around it, for example, the configuration or motion.

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3 years ago

A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the

Dima020 [189]

Explanation:

Given that,

Velocity of the proton in lab frame $u_{x} = 0.6c$

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

$u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}$

$u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}$

$u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})$

$u'=0.385c$

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)$

Where, Proton mass energy = m₀c²

Put the value into the formula

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)$

$K.E=234.57\ MeV$

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)$

$K.E=78.366\ MeV$

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

$P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}$

We know that,

Proton mass energy = m₀c²

$m_{0}=\dfrac{938.28}{c^2}$

$P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}$

$P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}$

$P_{obs}=150.69\ MeV/c$

Hence, This is required solution.

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3 years ago

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Answer:

Explained below:

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o-na [289]

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3 years ago

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