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Alexus [3.1K]
3 years ago
15

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

sthe current in the primary?and WHy?
Physics
1 answer:
Neporo4naja [7]3 years ago
8 0

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

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A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6
Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6  = V/0.0001

V = 2.0 × 10^6  * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q =  1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0
2 years ago
A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener
trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

U=mgh

where

m is its mass

g=9.8 m/s^2 is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

U=(m)(9.8)(15)=147 m (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

U=K=\frac{1}{2}mv^2

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

147 m = \frac{1}{2}mv^2

And re-arranging for v, we find the velocity:

v=\sqrt{2\cdot 147}=17.1 m/s

5 0
3 years ago
By the term universe astronomers mean
OverLord2011 [107]

'Universe' means 'Everything'. That is, all matter, all space, all time.

6 0
3 years ago
Suppose the polar bear was running on land instead of swimming. If the polar bear runs at a speed of about 8.3 m/s, how far will
olga_2 [115]

Answer:

298800 m

Explanation:

v =  \frac{d}{t}

V=speed

d=distance

t=time

d = vt

Change hours to seconds

It will be 36000s

8.3m/s * 36000s

=298800m

8 0
3 years ago
Gauss's law: Group of answer choices can always be used to calculate the electric field. relates the electric field throughout s
kvasek [131]

Answer:

relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss Law states that overall electric flux of a closed surface is equivalent right to charge enclosed which is divided by the permittivity. In other words Gauss Law stress that

net electric flux that pass through an hypothetical closed surface is equivalent to overall electric charge present within that closed surface.

The Gauss law can be expressed mathematically as

ϕ = (Q/ϵ0)

Q = total charge within the surface,

ε0 = the electric constant

5 0
3 years ago
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