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3 years ago

Which type of energy is the original source for the energy that food molecules can provide?

Physics

1 answer:

Svet_ta [14]3 years ago

4 0

Answer:

Chemical potential energy

Explanation:

which is used to for any form of energy.

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1. An astronaut in a spacesuit has a mass of 80 kilograms. What is the weight of this astronaut on the surface of the Moon where

Andrews [41]

Weight = 80 x (9.8/6) = .... N

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2 years ago

Key concepts <br><br> (Example | scientific method

nekit [7.7K]

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3 years ago

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Part K Find the y-component of the stone's velocity when it is 8.00 m below your hand

Romashka-Z-Leto [24]

The y-component of the stone's velocity when it is 8 m below the hand is 14.86 m / s

v² = u² + 2 a s

s = Displacement

u = Initial velocity

a = Acceleration

u = 8 m / s

s = 8 m

v² = 8² + 2 * 9.8 * 8

v² = 64 + 156.8

v = √ 220.8

v = 14.86 m / s

The equation used to solve the problem is an equation of motion. These equations are designed to locate an object in motion using components such as velocity, displacement, acceleration and time.

Therefore, the y-component of the stone's velocity is 14.86 m / s

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1 year ago

Which group of people was most directly affected by the twenty sixth amendment

vladimir2022 [97]

Adults. 18years olds and older were affected because back then only 21 years olds and older could vote but the 26th amendment changed that

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3 years ago

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A model rocket is launched straight upward with an initial speed of 56.5 m/s. It accelerates with a constant upward acceleration

marta [7]

Answer:

Maximum height reached by the rocket, h = 202.62 meters

Explanation:

It is given that,

Initial speed of the model rocket, u = 56.5 m/s

Constant upward acceleration, $a=1.96\ m/s^2$

Distance traveled by the engine until it stops, d = 198.8 m

Let v is the speed of the rocket when the engine stops. It can be calculated using the third equation of motion as :

$v=\sqrt{u^2+2ad}$

$v=\sqrt{(56.5)^2+2\times 1.96\times 198.8}$

v = 63.02 m/s

At the maximum height, v = 0 and the engine now decelerate under the action of gravity, a = -g. Let h is the maximum height reached by the rocket.

Again using third equation of motion as :

$v^2-u^2=-2gh$

$h=\dfrac{v^2-u^2}{-2g}$

$h=\dfrac{u^2}{2g}$

$h=\dfrac{(63.02)^2}{2\times 9.8}$

h = 202.62 meters

So, the maximum height reached by the rocket is 202.62 meters. Hence, this is the required solution.

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3 years ago