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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:

Oksana_A [137]3 years ago

8 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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BIO EST The archerfish (genus Toxotes) preys on land-based insects inhabiting tree limbs above the water. It has the ability to

monitta

Answer:

The initial speed of the water droplets is approximately 14.904 meters per second.

Explanation:

We can describe the water jet from archerfish as part of parabolic motion, which consists of the superposition of two different motions. First, an horizontal motion at constant velocity and, second, a free fall motion. The maximum height is reached when vertical component of speed is zero. The equation of motion is described below:

$v^{2} = v_{o}^{2}\cdot \sin^{2} \theta + 2\cdot g \cdot (y-y_{o})$

Where:

$v_{o}$ - Initial speed of the water jet, measured in meters per second.

$v$ - Current speed of the water jet, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{o}$ - Initial height, measured in meters.

$y$ - Current height, measured in meters.

Now we clear the initial speed within equation:

$v_{o}^{2} =\frac{v^{2}-2\cdot g\cdot (y-y_{o})}{\sin^{2}\theta}$

$v_{o} = \sqrt{\frac{v^{2}-2\cdot g\cdot (y-y_{o})}{\sin^{2}\theta} }$

If we know that $v = 0\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ , $y = 10\,m$ , $y=0\,m$ and $\theta = 70^{\circ}$ , the initial speed of the water droplets is:

$v_{o} = \sqrt{\frac{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (10\,m-0\,m)}{\sin^{2}70^{\circ}} }$

$v_{o} \approx 14.904\,\frac{m}{s}$

The initial speed of the water droplets is approximately 14.904 meters per second.

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3 years ago

Three ropes A, B and C are tied together in one single knot K. (See figure.)

masha68 [24]

The tension in the rope B is determined as 10.9 N.

<h3>Vertical angle of cable B</h3>

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

<h3>Angle between B and C</h3>

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

Learn more about tension here: brainly.com/question/24994188

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Answer:

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<em>The answer you are looking for is: </em>
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<em>Hope that helps!! </em>
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