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Iteru [2.4K]
3 years ago
15

An athlete runs a 100 m race in 10 seconds against a friction force of 100N. How do I work out his power output?

Physics
1 answer:
Rashid [163]3 years ago
6 0
They seem to cancel each other out which is odd
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The effective acceleration or deceleration due to gravity depends on the inclined angle of the track relative to ground; the steeper the slope is the greater the effective acceleration.
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What can happen to solar radiation when it enters atmosphere​
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The solar radiation is first intercepted by Earth's atmosphere, just a small part of the radiation is absorbed by gases such as water vapor. Some of the radiation is reflected back to space by the clouds and Earth's surface.

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Is hydrogen balloon element mixture or compound
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compound

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5 0
2 years ago
A rope exerts a 35N force on an object at an angle of 12N degrees above the horizontal. What horizontal and vertical components
nekit [7.7K]

Answer:

The horizontal component of the force, F_x= 34.24 \ N

The  vertical component of the force, F_y=7.28 \ N

Explanation:

Given;

Force on the rope, F = 35 N

angle between the rope and the horizontal = 12 °

The horizontal component of the force is given by;

F_x = Fcos \theta\\\\F_x = 35cos(12^0)\\\\F_x = 34.24 \ N

The vertical component of the force is given by;

F_y = Fsin\theta\\\\F_y = 35sin(12^0)\\\\F_y = 7.28 \ N

5 0
3 years ago
There are roughly 3 × 10−6 g of uranium per liter of seawater. Assuming there are 3.535 × 1020 gallons of seawater on earth, cal
Anvisha [2.4K]

Answer:

4 x 10^12 kg Uranium

Explanation:

As we know that 1 gallon = 3.785 L

Amount of Uranium in 1 L sea water = 3 x 10^-6 g

3.535 x 10^20 gallons = 3.535 x 10^20 x 3.785 L = 13.38 x 10^20 L

1 L of sea water contains =  3 x 10^-6 g Uranium

13.38 x 10^20 L of sea water contains = 13.38 x 10^20 x 3 x 10^-6 g Uranium

                                                              = 4 x 10^15 g = 4 x 10^12 kg Uranium

3 0
3 years ago
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