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2 years ago

5

The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

ope is 9.21 × 104 N, and the mass of Earth is 5.98 × 1024 kg, what is the mass of the telescope? Round the answer to the nearest whole number

2 answers:

Luda [366]2 years ago

5 0

The answer is 11,121 kg

PtichkaEL [24]2 years ago

4 0

The gravitational attraction between the Earth and the Hubble telescope is given by:

$F=G\frac{M m}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

m is the Hubble mass

r is the distance of the Hubble telescope from the earth's center

By rearranging the equation and substituting the numbers given by the problem, we find the mass of the telescope:

$m=\frac{F r^2}{GM}=\frac{(9.21 \cdot 10^4 N)(6.94 \cdot 10^6 m)^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(5.98 \cdot 10^{24}kg) } = 11121 kg$

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Answer:

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Explanation:

Weight = 700N

$F= m*g*h \\m*g= 700N\\F= 700N*h\\F=700N*8m\\F= 5600 N*m \\F=5600 J$

The power is the work in (Joules) or (N*m) in a determinate time (s) to get Watts (W) units for work

$Work= \frac{5600 J}{10s} = \frac{5600 \frac{kg*m^{2} }{s^{2} } }{10 s} \\Work= 560\frac{kg*m^{2} }{s^{3} } \\Work =560 W$

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2 years ago

Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer

SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

b

The efficiency is 99.4%

Explanation:

From the question we are given are given that

The transformer has Mega Volt Amp rating of 25MVA

The frequency is 60-Hz

Voltage rating 8.0kV : 78kV

The short circuit test gives : 453kV,321A,77.5kW

The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

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Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have

$\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}$

The volt will cancel each other

$\frac{25*10^6}{8*10^3} = 3125\ A$

Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

$MVA \ rating = (86*10^3)(3125) =268.75$

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil $I_2$ as shown in the circuit diagram can be obtained as

$I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321$

Now the efficiency can be obtained as thus

$\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}$

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ohaa [14]

Answer:

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Explanation:

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