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3 years ago

Can charging by friction occur only in solids? explain using an example.

1 answer:

8 0

The answer to the question can charging by friction occur only in solids is no. not at all. the static charge generated by following fluids can be hazardous. just as the static discharge when you touch a doorknob produces a spark, lightening is the spark from the static charge built up in the atmosphere.

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True or False

Ann [662]

The is true because I said so

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2 years ago

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Which is a characteristic of all ions? They are made of one type of atom. They have one overall charge. They are made of two or

Ira Lisetskai [31]

Answer:

They are made of one type of atom.

Explanation:

8 0

3 years ago

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How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns

Tasya [4]

Complete Question

How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.

The output from the secondary coil is 12 V

Answer:

The value is $N_s = 21 \ turns$

Explanation:

From the equation we are told that

The input voltage is $V_{in} = 120 \ V$

The number of turns of the primary coil is $N_p = 210 \ turn$

The output from the secondary is $V_o = 12V$

From the transformer equation

$\frac{N_p}{V_{in}} =\frac{N_s}{V_o}$

Here $N_s$ is the number of turns in the secondary coil

=> $N_s = \frac{N_p}{V_{in}} * V_s$

=> $N_s = \frac{210}{120} * 12$

=> $N_s = 21 \ turns$

4 0

3 years ago

A firefighter of mass 81 kg slides down a vertical pole with an acceleration of 3 m/s 2 . The acceleration of gravity is 10 m/s

natima [27]

Answer:

The force of friction that acts on him is

$F_k=567N$

Explanation:

The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:

$F=m*a$

$F=F_w-F_k$

$m*a=F_w-F_k$

$F_w=81kg*10m/s^2=810N$

Sole to Fk

$81kg*3m/s^2=810N-F_k$

$F_k=810N-243N$

$F_k=567N$

4 0

2 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

4 0

3 years ago