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3 years ago

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A plastic rod and a piece of cloth are both uncharged. A student rubs the plastic rod with the cloth. The plastic rod becomes ne

gatively charged. Compared with the plastic rod, which row of the table is correct for the charge on the cloth?

1 answer:

Molodets [167]3 years ago

7 0

Answer: a sock and a Pepe when the Pepe is rubbed by the sock the energy is enormous

Explanation:

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A circular coil of 124 turns has a radius of 1.86 cm. (a) Calculate the current that results in a magnetic dipole moment of magn

Blababa [14]

Answer:

(a) 24.35 A

(b) 0.089 Nm

Explanation:

N = 124

R = 1.86 cm = 0.0186 m

M = 3.28 A m^2

(a)

M = N I A

3.28 = 124 x I x 3.14 x 0.0186 x 0.0186

I = 24.35 A

(b) For maximum torque, angle = 90 degree

τ = M B Sin 90

τ = 3.28 x 27.4 x 10^-3 = 0.089 Nm

8 0

3 years ago

PLEASE ANSWER THESE I NEED HELP!!!

Gnesinka [82]

4. Intermediate conductivity and a low melting point.

Hope I helped :) mark as brainliest**

4 0

3 years ago

Read 2 more answers

A block of mass 500g is pulled from rest on ahorizontal frictionless bench by a steady force F and travels 8m in 2s find the acc

Korvikt [17]

<u>We are Given:</u>

Mass of the block (m) = 500 grams or 0.5 Kg

Initial velocity of the block (u) = 0 m/s

Distance travelled by the block (s) = 8 m

Time taken to cover 8 m (t)= 2 seconds

Acceleration of the block (a) = a m/s²

<u>Solving for the acceleration:</u>

From the seconds equation of motion:

s = ut + 1/2* (at²)

<em>replacing the variables</em>

8 = (0)(2) + 1/2(a)(2)²

8 = 2a

a = 4 m/s²

Therefore, the acceleration of the block is 4 m/s²

3 0

3 years ago

At a certain altitude above the Earth's surface, the electric field has a magnitude of 135 V/m. How much energy is stored in 1.0

Paul [167]

Answer:

Explanation:

Given that,

Electric field E=135V/m

Energy stored in 1m³of air=?

The energy stored in an electric field is given as

u = ½ εo E²

Where

U is the energy stored

εo is permissivity and it value is 8.85×10^-12C²/N..m²

And E is the electric field

Then,

U=½×8.85×10^-12×135²

U=8.06×10^-8J/m³

Then, the energy stored in 1m³ of air is 8.06×10^-8 J/m³

4 0

3 years ago

A skateboarder is moving at a constant speed of

Flauer [41]

The amount of time that passes from when this skateboarder begins to slow down until she begins to move back down the incline is 8.75 seconds.

<u>Given the following data:</u>

Speed = 1.75 m/s

Acceleration = -0.2 $m/s^2$

To calculate the amount of time that passes from when this skateboarder begins to slow down until she begins to move back down the incline, we would apply the first equation of motion;

<h3>The formula for the first equation of motion.</h3>

Mathematically, the first equation of motion for a decelerating object is given by this formula;

$V=U-at$

<u>Where:</u>

a is the acceleration.

V is the final speed.

U is the initial speed.

t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

$1.75 = 0-(-0.2)t\\\\1.75=0.2t\\\\t=\frac{1.75}{0.2}$

Time, t = 8.75 seconds.

Read more on acceleration here: brainly.com/question/24728358

8 0

3 years ago