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3 years ago

A 2-C charge experiences a force of 40 N when put at a certain location in

Physics

1 answer:

Pani-rosa [81]3 years ago

3 0

Answer:

E = 20 N/C

Explanation:

Given that,

Charge, q = 2 C

Force experience, F = 40 N

We need to find the electric field at that location.

The electric field in terms of electric force is given by :

F = qE

Where

E is the electric field

$E=\dfrac{F}{q}\\\\E=\dfrac{40\ N}{2\ C}\\E=20\ N/C$

So, the electric field at that location is 20 N/C.

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The total momentum of two marbles before a collision is 0.06 kg-m / s. no outside forces act on the marbles. what is the total m

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3 years ago

Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters.

erma4kov [3.2K]

Answer:

The extent to which it would stretch is $\Delta L = 0.015 \ m$

Explanation:

From the question we are told that

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The tension is $T =1500 N$

The Young modulus is mathematically represented as

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Where $\sigma$ is the stress which is mathematically represented as

$\sigma = \frac{F}{A}$

Substituting values

$\sigma = \frac{1500}{0.500*10^{-6}}$

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And e is the strain which is mathematically represented as

$e = \frac{\Delta L}{L }$

Where $\Delta L$ The extension of the steel string

Substituting these into the equation above

$Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

Substituting values

$2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

$\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}$

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3 years ago

Suppose that the metal cylinder in the last problem has a mass of 0.10 kg and the coefficient of static friction between the sur

andrew11 [14]

Answer:

The speed maximum speed is $0.49ms^{-1}$

Explanation:

The centrifugal force always acts on the cylinder and move away the rotating platform from the rotational axis. so the centripetal force provide by the frictional force:

Therefore

$\frac{mv^{2} }{r} =u_{s} mg$

coefficient of static friction: $u_{s} =0.12$

mass of the cylinder: $m=0.10kg$ \

distance of the cylinder from the turntable: $r=0.20m$

$\frac{mv^{2} }{r} =u_{s} mg$

cross multiply to find v

$v^{2} =u_{s} rg\\v=\sqrt{u_{s}rg } \\v=\sqrt{0.12*0.20*9.80}\\ v=0.49m/s$

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3 years ago

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