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Elden [556K]
3 years ago
8

How many excess electrons are on a ball with a charge of -1.4944 10-16 C?

Physics
1 answer:
agasfer [191]3 years ago
7 0

Answer:

There are 934 excess electrons on the ball with charge -1.4944*10^(-16).

Step-by-Step Explanation:

A ball will usually be neutral and have no charge on it.

If it is negatively charged, it means there are excess electrons on it.

If we know the charge on the ball, we can count the excess no. of electrons.

We know that the charge on the ball = c = -1.4944*10^-16

Charge of one electron = e = 1.6*10^-19

Charge of n electrons on the ball = ne = c

⇒ n = c/e

n = (-1.4944*10^-16)/(1.6*10^-19)

n = 934 electrons

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Answer:

Mass of the disk will be 2.976 kg

Explanation:

We have given force F = 45 N

Radius of the disk r = 0.12 m

Angular acceleration \alpha =140rad/sec^2

We know that torque \tau =I\alpha

And \tau =Fr

So Fr=I\alpha , here I is moment of inertia

So 50\times 0.12=I\times 140

I=0.0428kgm^2

We know that moment of inertia I=\frac{1}{2}mr^2

So 0.0428=\frac{1}{2}\times m\times 0.12^2

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6 0
3 years ago
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Answer:

Solid-state

Explanation:

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Some examples of a solid-state device are light emitting diodes (LED), integrated circuit (IC), Transistors, liquid crystal display (LCD) etc.

A solid-state device such as a transistor, refers to a semiconductor component that is used to control the flow of voltage or current and as a gate (switch) for electronic signals. Thus, a transistor allows for the amplification, control and generation of electronic signals in a circuit.

Hence, solid-state devices need constant power to operate. The timing functions are initiated by the presence or absence of a separate "trigger" signal.

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6 0
3 years ago
A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat
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Answer:

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

Explanation:

As we know that there is no external torque on the system of two disc

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So we will have

L_i = L_f

now we have

L_i = (\frac{1}{2}MR^2)\omega_o

also we have

L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now from above equation we have

(\frac{1}{2}MR^2)\omega_o  = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now we have

\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

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