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aniked [119]
3 years ago
13

A 50kg hiker is standing on the edge of a cliff. Find the hiker’s gravitational potential energy if the cliff is 30m high. (Show

the equation, show your work and answer with units.)
Physics
2 answers:
DENIUS [597]3 years ago
4 0

Answer:

The answer to your question is:      Ep = 14715 Joules

Explanation:

Data

mass = 50 kg

Potential energy = ?

height = 30 m

g = 9.81 m/s²

Equation

                Ep = mgh

Substitution

                     Ep = (50)(9.81)(30)

                     Ep = 14715 Joules

CaHeK987 [17]3 years ago
3 0

Answer:

14715 joules

Explanation:

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Answer:

if it is not moving it is zero, like all things that aren't in motion

Explanation:

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2 years ago
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An airplane travels 80 m/s as it makes a horizontal circular turn which has a 0.80-km radius. What is the magnitude of the resul
Nadya [2.5K]

Answer: 600N

Explanation:

Centripetal force is the force that causes a body to move in a circular path.

Centripetal force = MV²/r

M = 75kg v = 80m/s r = 0.80km = 800m

Substituting the values given in the formula;

F = 75 × 80²/800

F = 600N

The magnitude of the resultant force on the 75kg pilot is 600N.

6 0
3 years ago
46 points :)
IgorC [24]

Answer:

Its is dividing by 2

Explanation:

It starts with 100 them it goes to 50, 25, 12.5 so its a cycle of dividing by 2

3 0
3 years ago
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Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is
saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

v_{i}: is the initial velocity = 20 m/s

A_{i}: is the inlet area of the nozzle = 60 cm²  

\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

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Answer:

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