What does the c in conptt stand for

A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what i

Neporo4naja [7]

Answer:

The correct solution will be "12.0 A".

Explanation:

The given values are:

$N_p= 500 \ turn$

$N_s= 200 \ turn$

$I_s= 3.0 \ A$

By using the transformer formula, we get

⇒ $\frac{N_p}{N_s} =\frac{I_s}{I_p}$

⇒ $I_p = I_s\times \frac{N_s}{N_p}$

On substituting the given values, we get

⇒ $=3.0 \ A\times \frac{2000}{500}$

⇒ $=12.0 \ A$

8 0

2 years ago

1 Write 7 next to the statements that are true.

Grace [21]

Answer: The pressure in a liquid dec reaches with depth. F

The pressure in a liquid increases with depth.

The upthrust on an object is larger when it is deeper in a pool. 7

The bottom of a dam is thinner than the top of a dam. F

The bottom of a dam is thicker than the top of a dam.

The pressure is bigger at the bottom of a lake because of the weight of water above it. 7

I think these are the answers.

7 0

2 years ago

What does muttered mean

Tema [17]

Muttered is the same thing like mumble. It's like when you are speaking and no one couldn't hear what you said. So Muttered is when you say something low .

3 0

2 years ago

Read 2 more answers

In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t

galina1969 [7]

Answer:

$W_x = 0.9156\ c$

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

$W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}$

u_x = 0.543 c

v_x = - 0.741 c

$W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}$

$W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}$

$W_x = \dfrac{1.284c}{1+0.402363}$

$W_x = 0.9156\ c$

Relative velocity of the particle is $W_x = 0.9156\ c$

5 0

2 years ago

An airplane during departure has a constant acceleration of 3 m / s².

Rama09 [41]

Constant acceleration of plane = 3m/s²

a) Speed of the plane after 4s

Acceleration = speed/time

3m/s² = speed/4s

S = 12m/s

The speed of the plane after 4s is 12m/s.

b) Flight point will be termed as the point the plane got initial speed, u, 20m/s

Find speed after 8s, v

a = 3m/s²

from,

a = v - u

t

3 = v - 20

8

24 = v - 20

v = 44m/s

After 8s the plane would've 44m/s speed.

6 0

2 years ago