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asambeis [7]
3 years ago
13

13. A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V a

nd the secondary is connected to a floodlight that draws 5A, what is the power output? Please show ALL of your work.
Physics
1 answer:
Lisa [10]3 years ago
8 0

Answer:

The output power is 2 kW

Explanation:

It is given that,

Number of turns in primary coil, N_p=250

Number of turns in secondary coil, N_s=500

Voltage of primary coil, V_p=200\ V

Current drawn from secondary coil, I_s=5\ A

We need to find the power output. It is equal to the product of voltage and current. Firstly, we will find the voltage of secondary coil as :

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

\dfrac{250}{500}=\dfrac{200}{V_s}

V_s=400\ V

So, the power output is :

P_s=V_s\times I_s

P_s=400\ V\times 5\ A

P_s=2000\ watts

or

P_s=2\ kW

So, the output power is 2 kW. Hence, this is the required solution.

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Her mass is 65 kilograms
5 0
3 years ago
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A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at
defon

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been  ( To / (√ (1 - (v²/c²))).  where To = 2.00 sec

8 0
3 years ago
A snowball is launched horizontally from the top of a building at v = 19.6 m/s. If it lands d = 64 meters from the bottom, how h
tiny-mole [99]

Answer: height of building = 18.8m

Explanation: The question is a projectile motion, a two dimensional motion with a vertical constant acceleration (g = - 9.8m/s²) and a constant horizontal velocity (thus making horizontal component of acceleration zero).

From the question, distance between bottom of building and where the object lands = 64m, initial velocity for throwing the object = 19.6m/s

The horizontal range formulae is given as

d= vt

Where d= horizontal range = 64m, v = initial velocity of throw.

64 = 19.6 × t

t = 64/ 19.6

t = 3.265 s.

Height (h) of the building is gotten by using the formulae

h =vt - 1/2gt²

h = (19.6×3.265) - 1/2×9.8×(3.265)²

h = 71.05 - (104.47/2)

h = 71.05 - 52.235

h = 18.8m

6 0
3 years ago
. A driver travels 4.25 km. in two minutes. If the driver’s initial velocity is 20m/s, what is the acceleration?
Novosadov [1.4K]

Answer:

32.9166667 m / s^2

Explanation:

s = 4.25km (1000m / 1km)

= 4250m

u = 20m/s

delta T = 20min (60sec / 1min)

= 1200s

Use formula s = ut + (1/2)at^2

4250m = 20m/s * 1200s + (1/2)a*1200s^2

Rearrange it to find a

a = (s-ut)  / (1/2 * t^2)

a = (4250m - 20m/s*1200s) / (1/2 * 1200s^2)

a = -32.9166667 m / s^2

4 0
3 years ago
two objects were lifted by machine one object had a massive 2 kg it was lifted at a speed of 2 m/s the other had a massive 4 kg
rosijanka [135]
Kinetic energy =(1/2) (mass) (speed²)

First object:  KE = (1/2) (2 kg) (2m/s)² =  4 joules during the lift.

Second object:  KE = (1/2) (4kg) (3 m/s)² = 18 joules during the lift.

The second object has more kinetic energy while it's being lifted
than the first object has while it's being lifted. Once they reach their
final heights and stop, neither object has any kinetic energy. 
8 0
4 years ago
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