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3 years ago

What magnitude charge creates a 32.08 N/C electric field at a point 2.84 m away?

Physics

1 answer:

sammy [17]3 years ago

5 0

Answer:

Q = 29.4 x 10⁻⁹ C.

Explanation:

Electric field due to a charge Q at distance d is given by coulomb law as follows

Electric field

E = k Q /d²

where for air k which is a constant is 9 x 10⁹

$E=\frac{k\times Q}{d^2}$

Given E = 32.08 , d = 2.84 m

Putting these values in the relation above, we have

$[tex]32.08=\frac{9\times 10^9\times Q}{(2.84)^2}$ [/tex]

Q = 29.4 x 10⁻⁹ C.

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3 years ago

A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the

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To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

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3 years ago

A car with a mass of 2.0 * 10^3 kg is traveling at 15 m/s. what is the momentum of the car?

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Hello,

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3 years ago

a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her

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3 years ago

Read 2 more answers

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$

Substitute the given values

$u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56$

6 0

3 years ago