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3 years ago

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A generator produces 60 A of current at 120 V. The voltage is usually stepped up to 4500 V by a transformer and transmitted thro

ugh a power line of total resistance 1.0 Ω. Find the percentage power lost in the transmission line if the voltage is not stepped up.

1 answer:

aalyn [17]3 years ago

5 0

Answer:

The percentage power lost in the transmission line if the voltage not stepped up is 50%.

Explanation:

Given that,

Current = 60 A

Voltage = 120 V

Resistance = 1.0 ohm

We need to calculate the power

Using formula of power

$P=I\times V$

Where,I =current

V = voltage

Put the value into the formula

$P=60\times120$

$P=7200\ W$

We need to calculate the percentage power lost in the transmission line

If the voltage is not stepped up

Then, the power loss

$P'=I^2\times R$

Put the value into the formula

$P'=(60)^2\times1$

$P'=3600\ W$

The percentage power loss P''

$P''=\dfrac{P'}{P}\times100=\dfrac{3600}{7200}\times100$

$P''=50\%$

Hence, The percentage power lost in the transmission line if the voltage not stepped up is 50%.

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A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns throu

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Answer:

The time interval is $t = 3 \ s$

Explanation:

From the question we are told that

The angular acceleration is $\alpha = 4.0 \ rad/s^2$

The time taken is $t = 4.0 \ s$

The angular displacement is $\theta = 80 \ radians$

The angular displacement can be represented by the second equation of motion as shown below

$\theta = w_i t + \frac{1}{2} \alpha t^2$

where $w_i$ is the initial velocity at the start of the 4 second interval

So substituting values

$80 = w_i * 4 + 0.5 * 4.0 * (4^2)$

=> $w_i = 12 \ rad/s$

Now considering this motion starting from the start point (that is rest ) we have

$w__{4.0 }} = w__{0}} + \alpha * t$

Where $w__{0}}$ is the angular velocity at rest which is zero and $w__{4}}$ is the angular velocity after 4.0 second which is calculated as 12 rad/s s

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3 years ago

What is the frequency of radiation whose wavelength is 11.5 a0 ?

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Answer:

The frequency of radiation is $2.61 \times 10^{17} s^{-1}$

Explanation:

Given:

Wavelength $\lambda = 11.5 \times 10^{-10}$ m

Speed of light $c = 3 \times 10^{8}$ $\frac{m}{s}$

For finding the frequency of radiation,

$c = f \lambda$

$f = \frac{c}{\lambda}$

$f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }$

$f = 2.61 \times 10^{17}$ $s^{-1}$

Therefore, the frequency of radiation is $2.61 \times 10^{17} s^{-1}$

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3 years ago

A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after

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Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

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