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algol13
3 years ago
11

The transformer supplying power to an artist's light sculpture provides 9600-V AC when supplied by 120-V AC. If there are 100 tu

rns in the transformer's primary coil, how many turns are there in its secondary coil
Physics
1 answer:
jarptica [38.1K]3 years ago
5 0

Answer:

1.25 turns

Explanation:

From Transformer equation,

Ns/Np = Vs/Vp............................ Equation 1

Where Ns = Secondary turns, Np = Primary turns, Vs = secondary Voltage, Vp = Primary Voltage.

make Ns the subject of the equation  1

Ns = NpVs/Vp.......................... Equation 2

Given: Np = 100 turns, Vp = 120 V, Vs = 9600 V

Substitute into equation 2

Ns = 100(120)/9600

Ns = 1.25  turns.

Hence the are 1.25 turns in the  secondary coil = 1.25 turns

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Supply the missing force necessary to achieve equilibrium. Show your work.
Mumz [18]

<u>Analysing the Question:</u>

We know that equilibrium is the state of a body when it has equal and opposite forces being applied on it

In this case, a net downward force of 496N is being applied and a net upward force of (106 + 106 + 142 + x) N

<u>Finding the missing force:</u>

Since we have to achieve equilibrium, the net upward forces have to be equal to the net downward forces

So,  (106 + 106 + 142 + x) = 496

354 + x = 496

x = 496 - 354

x = 142 N

Therefore, the missing force is 142 N

8 0
3 years ago
A crate of mass 190 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, an
Oxana [17]

Answer:

Explanation:

Mass of 190kg

Coefficient of static friction is 0.4

Coefficient of kinetic friction 0.36

Horizontal force= 500N

Taking g=9.81m/s^2.

The weight of the body my

W=190×9.81=1863.91N

There is a normal acting on the body which is equal to the weight

N=W=1863.91N

Frictional force(fr) is acting on the body and it is opposite the horizontal force.

The minimum force to be overcome before the object can start to move is Fr = μsN

Fr= μsN. μs=0.4

Fr= 0.4×1863.91

Fr=745.56N.

Since the horizontal force (500N) is not up to the minimum force to make the object move, then the force of 500N the body is still at rest.

Then the frictional force at that time is equal to the horizontal force

Therefore

Functional force = 500N

b. Mass of asteroid is

M=2000kg

Asteroid velocity at a particular instant is,

U=(-1.30x10^4, 4.20x10^4, 0)m/s

Magnitude of U is

U=√(-1.30×10^4)^2 +(4.2×10^4)^2+0

U=√1.933E9

U=4.39×10^4m/s

Position of the asteroid from the centre of the earth is,

R= (6.00x10^6, 10.00x10^6, 0)m.

The magnitude of the radius is

R = √(6.00x10^6)^2+ (10.00x10^6)^2+ 0^2

R=√3.6E13+10E13+0

R=√13.6E13

R=1.17E7m

R^2=13.6E13m

The mass of the earth is

Me=5.97x10^24 kg

The momentum of the asteroid after time, t=1.5×10^3s

Given that G=6.67x10^-11Nm^2/kg^2

Momentum is

Mv-Mu=Ft

There the new momentum will be

Mv=Ft+Mu

Now we the to find the force the earth exert on the asteroid by using

F=GMMe/R^2

F=6.67E-11 ×2000× 5.97E24 /13.6E13

F=7.964E17/13.6E13

F=5855.88N

The new momentum

Mv= Mu+Ft

Mv= 2000(4.39E4)+5855.88(1.5E3)

Mv=9.66E7kgm/s

The new momentum is 9.66×10^7 Kgm/s

8 0
3 years ago
A long coaxial cable (Fig. 2.26) carries a uniform volume charge density rho on the inner cylinder (radius a), and a uniform sur
Yuki888 [10]

Answer:

a) E = ρ / e0

b) E = ρ*a / (e0 * r)

c) E = 0

Explanation:

Because of the geometry, the electric field lines will all have a radial direction.

Using Gauss law

Q/e0 = \int \int E * dA

Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.

We can make this cylinder of different radii to evaluate the electric field at different points.

Then:

A = 2*π*r (area of cylinder per unit of length)

Q/e0 = 2*π*r*E

E = Q / (2*π*e0*r)

Where Q is the charge contained inside the cylinder.

Inside the cable core:

There is a uniform charge density ρ

Q(r) = ρ * 2*π*r

Then

E = ρ * 2*π*r / (2*π*e0*r)

E = ρ / e0 (electric field is constant inside the charged cylinder.

Between ther inner cilinder and the tube:

Q = ρ * 2*π*a

E = ρ * 2*π*a / (2*π*e0*r)

E = ρ*a / (e0 * r)

Outside the tube, the charges of the core cancel each other.

E=0

4 0
3 years ago
A 75 g bullet is fired from a rifle having a barrel 0.650 m long. Assuming the origin is placed where the bullet begins to move,
kolbaska11 [484]

Answer:

7557.875 J

Explanation:

Assumimg we need to find work done by the bullet on the gas

The work is found by integrating force over distance

W= W=\int_{x_i}^{x_f}F.ds

For the force in this problem we have

W=\int_{0}^{0.65}14000 + 10000x - 21000x^2.ds

= [14000x+10000x^2-21000x^3]_{0}^{0.65}

=7557.875 J

8 0
3 years ago
What is the pressure level on surface of titan?
Tcecarenko [31]
<span>150 kPa
While Titan is Saturns largest moon, it only has a mass of 0.0225 Earths
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4 0
3 years ago
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