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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

Read 2 more answers

When beryllium-7 ions (m = 11.65 × 10-27 kg) pass through a mass spectrometer, a uniform magnetic field of 0.205 T curves their

AysviL [449]

Answer:

ratio =0.3075 T

Explanation:

The magnetic field B creates a force on a moving charge such that

$F = qvB$

Now this causes a centripetal acceleration

$F = = mv^2/r$

$qvB = mv^2/r$ ...........(i)

$B = mv/(rq)$ ...............(ii)

If accelerating potential V is same and then kinetic energy equals the potential energy difference

$\frac{1}{2} mv^2 = Vq$

$v = \sqrt{(2Vq/m)}$ put these value in equation (ii)

$B = m\frac{\sqrt{(2Vq/m)} }{rq}$

simplifying we get

$B =m \frac{(\sqrt{ 2Vm/q})}{r}$

for same location r will be same in both case

$B_{7} = \frac{ \sqrt{(m_{7})(2V/q) }}{r}$ ..............(iii)

$B_{10} = \frac{ \sqrt{(m_{10})(2V/q) }}{r}$ ..........(iv)

dividing (iv) and (iii) equation we get

$\frac{B_{10}}{B_{7}} = \sqrt{\frac{m_{10}}{{m_7}} }$

${B_{10}} = B_{7} \sqrt{\frac{m_{10}}{{m_7}} }$

$B_{10} = 0.2574T\sqrt{\frac{ (1.663x10^-26}{(1.165x10^-26)}$

so on solving we get

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3 years ago

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nevsk [136]

Answer:

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