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2 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]2 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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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

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2 years ago

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh

S_A_V [24]

Answer:

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Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

V₀² = 3.04 m/(0.10126 s²/m)

V₀ = √30.02 m²/s²

<u>V₀ = 5.47 m/s</u>

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2 years ago

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Answer:

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Explanation:

As we know that the position of maximum intensity on the screen is given as

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here we know that

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now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen

so we have

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so here we have

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$\lambda_2 = 573.3 nm$

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oksano4ka [1.4K]

Answer:

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Explanation:

Because that's how it works.

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