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zubka84 [21]
4 years ago
15

A step-down transformer providing electricity for a residential neighborhood has exactly 2750 turns in its primary. When the pot

ential difference across the primary is 5670 V, the potential difference at the secondary is 240 V. How many turns are in the secondary? Show all work.
Physics
1 answer:
zzz [600]4 years ago
5 0

Answer:

116

Explanation:

The transformer equation states that:

\frac{V_p}{V_s}=\frac{N_p}{N_s}

where

Vp is the voltage in the primary coil

Vs is the voltage in the secondary coil

Np is the number of turns in the primary coil

Ns is the number of turns in the secondary coil

Here we have:

Vp = 5670 V

Np = 2750

Vs = 240 V

So we can solve the formula to find Ns:

N_s = N_p \frac{V_s}{V_p}=(2750)\frac{240 V}{5670 V}\sim 116

So, the secondary coil has 116 turns.

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Identify the forces acting on the object of interest. From the list below, select the forces that act on the piano.
Nonamiya [84]

Answer:

gravitational force acting on the piano (piano's weight)

force of Chadwick on the piano

force of the floor on the piano (normal force)

Explanation:

Figure is missing: found it in attachment.

In the figure, we notice that the piano is accelerating along the horizontal direction: this means that there is a net force acting along this direction. This force is prodiced by Chadwick, and it acts in the same direction as the acceleration, so one force is:

force of Chadwick on the piano

Also, every object on Earth experencies the force of gravity, which is also called weight. The weight of the piano acts downward, so a second force is:

gravitational force acting on the piano (piano's weight)

Finally, we notice that the piano is in equilibrium along the vertical direction (no acceleration): this is because there is another force acting opposite to the piano's weight (and with equal magnitude), and this force is the normal force exerted by the floor on the piano:

force of the floor on the piano (normal force)

3 0
3 years ago
At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci
siniylev [52]

Answer:

r = 1.86 m

Explanation:

Here the force due to wall of the cylinder is towards the axis of the cylinder

This force will act as centripetal force for the people sit inside the chamber

now we will have

F_c = \frac{mv^2}{R}

now we will have

F_c = 488 N

m = 84.4 kg

v = 3.28 m/s

now we have

488 = \frac{84.4(3.28)^2}{r}

now we have

r = 1.86 m

8 0
3 years ago
A ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec. Assuming that the air resistance can
Volgvan

Answer:

Maximum height of the ball, h(t) = 27.56 m

Explanation:

It is given that, a ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec.      

The height of the ball as a function of time t is given by :

h(t)=h_o+v_ot-16t^2

h₀ is initial height, h₀ = 0

So, h(t)=42t-16t^2 .........(1)

For maximum/minimum height,  \dfrac{dh(t)}{dt}=0

42-32t=0...(2)

t = 1.31 s

Differentiating equation (2) wrt t

h''(t) = -32 < 0

So, at t = 1.31 seconds we will get the maximum height.

Put the value of t in equation (1)

h(t)=42\times 1.31-16\times (1.31)^2

h(t) = 27.56 m

Hence, this is the required solution.

7 0
4 years ago
In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radi
Vitek1552 [10]

Answer:

(a) 1,569.63 N

(b)  195,933.99 Pa

(c) As pressure and volume are equal for each piston, workdone must also be equal

(d) 1,647.47 kg

Explanation:

Let the cross-sectional  area (CSA) of small piston = A₁

Let the  cross-sectional  area (CSA) of the bigger piston  = A₂

Let the Force applied at the smaller piston  = F₁

Let the Force applied at the bigger piston  = F₂

The principle of hydraulic lift  assumes the that the fluid is in-compressible, resulting to a constant pressure system.

F₁/A₁ =F₂/A₂----------------------------------------------------------- (1)

(a)  F₁=  F₂ xA₁ /A₂

F₂  = 13,300 N

A₁ = π r₁²

    =π x (0.0505)²

   =  0.008011 m²

A₂= π r₂²

    =π x (0.147)²

   =  0.06788 m²

Substituting into (1)

F₁ = 13,300 x  0.008011/0.06788

   = 1,569.6272

   ≈ 1,569.63 N

(b)   Air pressure = Force/Area

                            =  F₁/A₁

                            = 1,569.6272/ 0.008011

                            =   195,933.99 Pa

(c) The pressure is constant for both pistons according to Pascal Law.

Workdone = force x distance----------------------------------------- (2)

force = pressure × area

distance = volume/area from

Substituting into (2)

Workdone = pressure × volume.

As pressure and volume are equal for each piston, work must also be equal

(d)  F₂  =  F₁ x  A₂/ A₁---------------------------------------------------- (3)

A₁ = π r₁²

    =π x (0.079)²

   =  0.01960 m²

A₂= π r₂²

    =π x (0.353)²

   =  0.3914 m²

Substituting into (2)

F₂= 825 x  0.3914/ 0.01960

   = 16,474.7448

   ≈ 16,474.74 N

  = 1,647.47 kg

 

3 0
3 years ago
Despite a very strong wind, a tennis player manages to hit a tennis ball with her racquet so that the ball passes over the net a
kobusy [5.1K]

Answer:

The downward force of gravity and the force exerted by the air.

Explanation:

Gravity, in a Newtonian Framework, acts at distance, so, the ball doesn't has to be connected to Earth to feel its gravitational pull (as we know by experience).

The racket only can exert a force in the ball when is touching it, so, once the ball has left the contact with the racket, there is no force by the "hit".

The air is always touching the ball on the atmosphere (even when physicist pretend that is not). So, there is drag from the wind, and a buoyant force  exerted by the air over the ball at every time.

5 0
4 years ago
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