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charle [14.2K]
2 years ago
7

How much work is done by a 120N force applied to a box that moves 5m?

Physics
1 answer:
Vsevolod [243]2 years ago
4 0

Answer:

600 Joules

Explanation:

Using the formula F*d*cosФ. Assuming the Ф is parallel to the motion. The work done is 600 Joules.

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If mass of both the objects are doubled
Fofino [41]

Answer:

it should be four times

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2 years ago
A 15-watt bulb is connected to a circuit that has a total of 60. Ω of resistance. How many electrons are passing through that bu
Mariulka [41]

Answer:

3.2075*10^16

Explanation:

Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.

3 0
2 years ago
A 42 Watt light bulb is connected to a 12 volt source of potential
abruzzese [7]

Answer:

The natural medium emanating from the Sun and other very hot sources (now recognised as electromagnetic radiation with a wavelength of 400-750 nm), within which vision is possible.

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just the way it is

8 0
3 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
How would playing a game of soccer, baseball, or basketball be different if inertia didn’t exist?
Elina [12.6K]
This would be a funny game.

Inertia, according to first Law of Newton, means that when you throw a ball it will continue moving unless a force acts over it.

Without inertia, the ball instead of continue moving, would stop as soon as it leaves your hand or foot. You had to stay pasted to the ball unitl it reaches the desired destiny for it continue moving. 
5 0
3 years ago
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