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frez [133]
3 years ago
8

The work done on a 10 kilogram mass to give it a speed of 5 meters per second is

Physics
1 answer:
mixas84 [53]3 years ago
7 0

Answer:

idrk

Explanation:

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AN airplane travels 4000 m in 20s on a heading 0f 35 degrees north west. Calculate average velocity .
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Isn't velocity Distance over time? if the degree isn't adding resistance it should be 4000 ÷ 20 which gives you 200mps ("per second") which is the velocity without resistance.
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Brainliest answer! You and a friend are at a park and want to swing on the swings. Describe when you would have the greatest pot
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A ray of yellow light ( f8= 5.09 × 1014 Hz) travels at a speed of 2.04×10 meters per second in
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The complete question in the attached figure
Let
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v ------------------- > is the speed in medium
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A box with a mass of 50 kg is raised straight up. What is the force of the box?
8_murik_8 [283]

Answer:

A box with a mass of 50 kg is raised straight up. What is the force of the box? plss help!! two types exist-positive and negative. possible answers: A- electric current B- repel C- lines of force D- charges. Five wavelengths are generated every 0.100 s in a tank of water.

Explanation:

I hope it helped

3 0
2 years ago
A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w
user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

p\propto T

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

\frac{p_1}{T_1}=\frac{p_2}{T_2}

where in this problem:

p_1=0.981 atm is the initial pressure of the gas

T_1=65^{\circ}+273=338 K is the initial absolute temperature of the gas

T_2=41^{\circ}+273=314 K is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm

3 0
4 years ago
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