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Solnce55 [7]
3 years ago
14

When a constant force acts upon an object, the acceleration of the object varies inversely with its mass 2kg. When a certain con

stant force acts upon an object with mass , the acceleration of the object is 26m/s^2 . If the same force acts upon another object whose mass is 13 , what is this object's acceleration
Physics
1 answer:
BARSIC [14]3 years ago
3 0

Answer:

The acceleration of the second object is 4 m/s².

Explanation:

Given that,

when the a constant force acts upon an object, the acceleration varies inversely with its mass .

mass\propto \frac{1}{acceleration}

We can rewrite the equation as

\frac{m_1}{m_2}=\frac{a_2}{a_1}

When an a certain constant force acts upon an object with mass 2 kg , the acceleration of the object is 26 m/s².

Here, m_1 = 2 kg, a_1 = 26 m/s²,m_2 = 13 kg and a_2=?

\frac{m_1}{m_2}=\frac{a_2}{a_1}

\frac{2}{13}=\frac{a_2}{26}

\Rightarrow \frac{a_2}{26}= \frac{2}{13}

\Rightarrow {a_2}= \frac{2}{13}\times 26

\Rightarrow a_2=4 m/s²

The acceleration of the second object is 4 m/s².

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Answer:

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(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

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when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

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F is the force between the charges

K is Coulomb's constant

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r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

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