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3 years ago

Mt. Everest is 20,028 feet high. How many miles is this? ( there are 5,280 feet in a mile)

Physics

2 answers:

MAXImum [283]3 years ago

6 0

20,028 feet is 3.793 miles

The summit of Mt. Everest is 29,029 ft above sea level. That's 5.498 miles.

Nana76 [90]3 years ago

3 0

<u>Answer:</u>

Mt. Everest is 3.79 miles high.

<u>Solution: </u>

Everest is 20,028 ft high.

As given, 1 mile = 5280 ft

So $1 \mathrm{ft}=\left(\frac{1}{5280}\right) \mathrm{mile}$

Then 20,028 ft will be $=20028 \times\left(\frac{1}{5280}\right) \text { mile }=\left(\frac{20028}{5280}\right) \text { mile }$

$=\left(\frac{10014}{2640}\right) \text { mile }=\frac{5007}{1320} \text { mile }=3.79 \text { mile }$

The height of Everest is 3.79 miles .

Note: The actual height of Mt. everest is 5.4979 miles or 29029 ft.

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Answer:

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Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

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3 years ago

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The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

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$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

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(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

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4 years ago

A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

almond37 [142]

Answer:

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Explanation:

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- The mass of motor = 4m

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What is the speed of the command module relative to Earth just after the separation?

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Initial momentum = Final momentum

Pi = Pf

M*vs_e = m*vc_e + 4m*vm_e

Where,

M = m + 4m = 5m

vc_e = Velocity of command relative to earth

vm_e = Velocity of motor relative to earth

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

vm_e = v_c/m + vc_e

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

5m*vs_e - 4m*(v_c/m) = 5m*vc_e

vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

vc_e = 3760 - 0.8*(90)

vc_e = 3,688 km/h

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3 years ago

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