Ask question

Ask question

All categories

schepotkina [342]

3 years ago

11

Hey I need help on this please???

1 answer:

Agata [3.3K]3 years ago

6 0

Answer: it's c or 4.0 m/s

Explanation:

You might be interested in

Why does the mass of a candle decrease after burning?

34kurt

Because when a candle burns the wax is burning off so as its heated more and more wax is lost and if more wax is lost... Obviously it means that the candles mass will be lower :)

3 0

4 years ago

What is paleomagnetism how has it been used to understand plate tectonics?

Nonamiya [84]

<span>Plate tectonics is the theory that the earth's crust is broken up into plates that float on top of a hotter and more fluid layer below. Evidence to support this theory has been uncovered through the study of the earth's past magnetic field, known as paleomagnetism.</span>

8 0

4 years ago

The difference in a brass rod and steel rod is same as 12cm at all temperature .Find the length of the rod at 0°C.

egoroff_w [7]

Answer:

I think its 12 cm itself. Because in the question it is said that the length is same in all temperature

5 0

3 years ago

What was the initial velocityof the t-shirt as it left the cannon?

anastassius [24]

Well nick, the velocity at time= 0 it’s zero meter/second.

5 0

3 years ago

I) A circular coil with radius 20 cm is placed with it's plane parallel and between two straight

Mamont248 [21]

Answer:

$B=15.433\ T$ inwards when viewing from the left side.

Explanation:

Given:

radius of the circular loop, $r=0.2\ m$

current in the coil, $I_c=0.5\ A$

Direction of current is clockwise when viewed from left.

Distance of wire P from the loop, $d_p=0.4\ m$

Distance of wire Q from the loop, $d_q=0.8\ m$

current in each wires P & Q, $I=0.2\ A$

Now the magnetic field in coil will be inwards when viewed from left by the Maxwell's right hand thumb rule.

<u>Magnitude is given by:</u>

$B_c=\frac{\mu_0.I_c}{2R}$

$B_c=\frac{4\pi\times 10^{-7}\times (0.5)}{2\times 0.2}$

$B_c=15.7\times 10^{-7}\ T$

<u>Now the effect of magnetic field due to wire P at the center of the loop:</u>

(We get the effective distance as 0.4+0.2=0.6 m)

$B_P=\frac{\mu_0.I}{2\pi.d_p}$

$B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 0.6}$

$B_P=6.67\times 10^{-8}\ T$ coming out of the loop when viewed from left.

<u>Now the effect of magnetic field due to wire Q at the center of the loop:</u>

(We get the effective distance as 0.8+0.2=1 m)

$B_P=\frac{\mu_0.I}{2\pi.d_q}$

$B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 1}$

$B_P=4\times 10^{-8}\ T$ going in to the loop when viewed from left.

<u>Now the net resultant effect all the magnetic fields:</u>

$B=B_c-B_P+B_Q$

$B=15.7-0.667+0.4$

$B=15.433\ T$ inwards when viewing from the left side.

7 0

3 years ago