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RoseWind [281]
3 years ago
6

5. An electrical power plant generates electricity with a current of 50 A and a potential difference of 20 000 V. In order to mi

nimize the power losses due to the high current heating up the line, a transformer steps up the potential difference to 500 000 V before it is transmitted. What is the current in the transmission lines?
Physics
1 answer:
lakkis [162]3 years ago
3 0

Answer: Current = 2 A

Explanation:

Given that an electrical power plant generates electricity with a

current I = 50 A

Potential difference V = 20 000 V

The resistance R will be achieved by Ohms law formula which state that

V = IR

But the power generated will be the product of potential difference and the current

Power P = IV

P = 50 × 20000

P = 1, 000000 W

When the transformer steps up the potential difference to 500 000 V before it is transmitted

Power is always constant.

Using the formula for power again with

V = 500000

1000000 = 500000× I

Make I the subject of formula

Current I = 1000000/500000

Current I = 2 A

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We can determine the velocity of a wave when given the frequency and the
Kobotan [32]

Hello.

The answer is: D. wavelength

This is correct because   frequency x wavelength = speed

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3 years ago
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A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i
Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

v^2=u^2+2as

0^2=14.5^2+2\times a\times 18.25

a=-5.76m/sec^2

We know that acceleration is given by

a=\mu g

5.76=\mu\times  9.81

\mu =0.587

So coefficient of friction will be 0.587

6 0
3 years ago
A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:
madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be 0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}. Accordingly, the ball's momentum at that moment would be p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right).

3 0
3 years ago
As the elephant falls from 10 m does it lose or gain KE? Explain.
ivolga24 [154]

He loss KE hope this helps

7 0
4 years ago
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A liquid is used to make a mercury-type barometer. The barometer is intended for space-faring astronauts. At the surface of the
Anarel [89]

Answer:

Density of liquid = 4730 kg/m³

Atmospheric pressure on planet X = 8401.7 N/m²

Explanation:

Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm

So, ρgh = ρ₁gh₁

ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³

The atmospheric pressure on planet X

P = ρg₁h₃     g₁ = g/4 and h₃ = 725 mm = 0.725 m

on planet X

P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²

6 0
4 years ago
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