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nikklg [1K]
3 years ago
15

a small negatively charged sphere with a mass of 5.4*10^-5 is suspended between two parallel plates. the potential difference is

360V to hold the sphere stationary. what is the sphere's charge?

Physics
1 answer:
labwork [276]3 years ago
3 0
Here, Fe = Fg
q.E = m.g
We have: E = 360 V
m = 5.4 × 10⁻⁵
g = 9.8 m/s²   [ constant value for earth system ]

Substitute their values into the expression:
q (360) = 5.4 × 10⁻⁵ × 9.8
q = 52.92 × 10⁻⁵ / 360
q = -1.47 × 10⁻⁶  [ negative sign represents the nature of charge ] 

So, Your Final answer would be 1.47 × 10⁻⁶

Hope this helps!
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madreJ [45]

Hello!

<em>a) What is the change in momentum</em>?

The change of the momentum is the velocity, because the velocity is reduced to zero.

We can calculate the aceleration, applicating the formula:

\boxed{a=\frac{V-Vi}{t} }

Like you see, the final velocity will be zero, so:

a = \dfrac{0m/s-40m/s}{4s}

a = \dfrac{-40m/s}{4s}

a = -10\ m/s^{2}

Like you see, the aceleration applicate by superman is of <u>-10 m/s^2.</u>

If the aceleration is negattive, means the velocity will decrease.

<em>b.) What is the magnitud of the force wich superman exerts on the train?</em>

For calculate the force applicate for superman, lets applicate the second law of Newton:

\boxed{F=ma}

Lets replace and resolve it:

F = 8x10^3 kg * (-10 m/s^2)

F = 8000 kg * (-10 m/s^2)

F = -80 000 N

The force applicated is of <u>-80000 Newtons.</u>

When the force is negative means the force applicated in the another direction of the object what is traveling.

8 0
3 years ago
A rock is thrown from a bridge at an angle 30∘ below horizontal.immediately after the rock is released, is the magnitude of its
Wittaler [7]
<span>The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.</span>
4 0
3 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
a 72kg person is standing on a bathroom scale (calibrated in newtons). what is the reading of the scale when the acceleration is
Likurg_2 [28]

Answer:

F=ma

therefore A=F/M

Explanation:

i think that's what your doing but I'm not sure

4 0
3 years ago
A parallel circuit contains an 18-V battery wired with 2 bulbs with resistances of 8
RSB [31]

Answer:

See below

Explanation:

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Equivalent resistance   =   1 / (1/8 + 1/24) = 6 Ω

7 0
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