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3 years ago

8

A centripetal force of 250 N is exerted on a 2.2 kg discus as it rotates uniformly in a horizontal circle of radius 0.85 m. Calc

ulate the speed of the discus in both m/s and rotations per minute?

1 answer:

Andreyy893 years ago

5 0

Answer:

133m/s

Explanation:

b/c 250÷2.2÷0.85=133

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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

a)

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

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Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the

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$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Heat capacity of body 1 :

$\qquad \sf \dashrightarrow \:m_1s_1$

Heat capacity of body 2 :

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$\qquad \sf \dashrightarrow \:m_1s_1 = m_2s_2$

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Now, consider specific heat of composite body be s'

According to given relation :

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[ since, $m_2s_2 = m_1s_1$ ]

$\qquad \sf \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}$

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$\qquad \sf \dashrightarrow \:s' = \dfrac{ 2 s_2}{ (\frac{s_2 + s_1}{s_1} )}$

$\qquad \sf \dashrightarrow \: s' = \dfrac{2s_1s_2}{s_1 + s_2}$

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