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3 years ago

A 06-C charge and a 07-C charge are apart at 3 m apart. What force attracts them?

Physics

1 answer:

Vlad [161]3 years ago

5 0

Answer:

Force of 37.8 × 10^(6) N attracts the two charges

Explanation:

The force between two charges is given by

F = k*q1*q2/r²

Where q1 and q2 are 0.06 C and 0.07 C.

r is the distance between q1 and q2 which is equal to 3 m

k is a constant = 9 × 10^(9) N.m²/C²

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

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When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

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3 years ago

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2 years ago

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Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

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3 years ago

A 06-C charge and a 07-C charge are apart at 3 m apart. What force attracts them?​

A 06-C charge and a 07-C charge are apart at 3 m apart. What force attracts them?