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xxMikexx [17]
3 years ago
7

Two small metal spheres are 25 cm apaft.The spheres have equal amount of negative charge and repel each other with a force of 0.

036 N. What is the magnitude of the charge on each sphere? (a) 2x10-3 C (h) 3 x 10-4C
Physics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

0.5\times 10^{-6}C

Explanation:

According to coulombs law force between two charges is given by F=\frac{1}{4\pi \epsilon _0}\frac{Q_1Q_2}{R^2} here R is the distance between both the charges which is given as 25 cm

We have given force F =0.036 N

So  0.036=\frac{1}{4\pi \times 8.85\times 10^{-12}}\frac{Q^2}{(0.25^2)} As \epsilon _0 is constant which value is 8.85\times 10^{-12}

Q^2=0.250\times 10^{-12}

Q=0.5\times 10^{-6}C

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A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.13 s. W
vitfil [10]

Answer:

Part a)

T = 0.52 s

Part b)

f = 1.92 Hz

Part c)

speed = 3.65 m/s

Explanation:

As we know that the particle move from its maximum displacement to its mean position in t = 0.13 s

so total time period of the particle is given as

T = 4\times 0.13 = 0.52 s

now we have

Part a)

T = time to complete one oscillation

so here it will move to and fro for one complete oscillation

so T = 0.52 s

Part b)

As we know that frequency and time period related to each other as

f = \frac{1}{T}

f = \frac{1}{0.52}

f = 1.92 Hz

Part c)

As we know that

wavelength = 1.9 m

frequency = 1.92 Hz

so wave speed is given as

speed = wavelength \times frequency

speed = 1.92 \times 1.9

speed = 3.65 m/s

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3 years ago
When you park your car uphill next to a curb, the right front wheel should be:?
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Certain rifles can fire a bullet with a speed of 950 m/s just as it leaves the muzzle (this speed is called the muzzle velocity)
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Answer:

a) By v^2 = u^2 + 2as => a= 70291.70.

(b)By v = u + at => t= 1.58 ms.

(c)By v^2 = u^2 - 2gh => H = 46045.92 m.

Explanation:

a) By v^2 = u^2 + 2as

(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70

(b)By v = u + at

950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms

(c)By v^2 = u^2 - 2gh

0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m

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An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

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2 years ago
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