Find the solid angle made by the part of surface area 2 cm square of square and radius 4 centimetre

Hurry need an answer

notsponge [240]

Answer:

i think b

Explanation:

6 0

3 years ago

The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. why then doesn't the heavier rock fall faster?

katrin2010 [14]

It takes twice the force to produce the same acceleration in the 2kg rock.

5 0

3 years ago

A 0.8 m length of wire is formed into a single turn, square loop in which there is a current of 12 A. The loop is placed in magn

rewona [7]

Answer:

0.073 N-m

Explanation:

i = 12 A, l = 0.8 m, B = 0.12 T

The circumference of the loop is 0.8 m.

Let r be the radius of the loop.

2 x 3.14 x r = 0.8

r = 0.127 m

Maximum Torque = i x A x B

Maximum Torque = 12 x 3.14 x 0.127 x 0.127 x 0.12 = 0.073 N-m

8 0

3 years ago

The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).

sammy [17]

Answer:

(a) $emf_L=-LI_{max}\omega cos(\omega t)$

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

$I(t)=I_{max}sin(\omega t)$ (1)

(a) The induced emf is given by the following formula

$emf_L=-L\frac{dI}{dt}$ (2)

You derivative the expression (1) in the expression (2):

$emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)$

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

$emf_L=-\omega LI_{max}$

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

6 0

3 years ago

Two charged spheres are 20 cm apart and exert an attractive force of 8 x 10-9 n on each other. What will the force of attraction

Pavel [41]

Answer:

$3.2\cdot 10^{-8} N$

Explanation:

The inital electrostatic force between the two spheres is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

$F=8\cdot 10^{-9} N$ is the initial force

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the two spheres

The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have

$r'=\frac{r}{2}$

Therefore, the new electrostatic force will be

$F'=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F$

So the force has increased by a factor 4. By using $F=8\cdot 10^{-9} N$ , we find

$F'=4(8\cdot 10^{-9} N)=3.2\cdot 10^{-8} N$

6 0

3 years ago

Find the solid angle made by the part of surface area 2 cm square of square and radius 4 centimetre​

Find the solid angle made by the part of surface area 2 cm square of square and radius 4 centimetre