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Masteriza [31]
3 years ago
9

A plane flying against a jet stream will travel faster than a plane traveling with a jet stream. Please select the best answer f

rom the choices provided T F
Physics
2 answers:
Yuliya22 [10]3 years ago
7 0

The simplified answer is, It is False

stepan [7]3 years ago
3 0

"with the wind" is a tail-wind, and the speeds are added to get the groundspeed.

"against the wind" is a head-wind, and the windspeed is subtracted from the airspeed.

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a 15 kg tv sots on a shelf at a height of 0.3 m. how much gravitational potential energy is added to the television when it is l
nata0808 [166]
Gravitational potential energy can be described as m*g*h (mass times gravity times height).

Originally,
15kg * 9.8m/s^2 *0.3 m = 44.1 kg*m^2/s^2 = 44.1 Joules.

After it is moved to a 1m shelf:
15kg * 9.8m/s * 1 = 147 kg*m^2/s^2= 147 Joules.

To find how much energy was added, we subtract final energy from initial energy:

147 J - 44.1 J = 102.9 Joules.
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How can a moving coil galvanometer can be made into a dc ammeter?
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3 years ago
What is not changed when work is done by a machine?
Irina18 [472]
B) The amount of work done
8 0
3 years ago
Read 2 more answers
Can at transform plate boundaries, two plate move toward each other true or false
nekit [7.7K]
 the answer to the question is false

7 0
2 years ago
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
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