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3 years ago

A stone of mass 0.2 kg falls with an acceleration of 10.0 m/s. How big is the force that causes this acceleration?

Physics

1 answer:

Kobotan [32]3 years ago

4 0

Answer:

$\boxed {\boxed {\sf 2 \ Newtons}}$

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F= m \times a$

The mass of the stone is 0.2 kilograms and the acceleration is 10.0 meters per square second.

m= 0.2 kg
a= 10.0 m/s²

Substitute the values into the formula.

$F= 0.2 \ kg * 10.0 \ m/s^2$

Multiply.

$F=2 \ kg*m/s^2$

Convert the units.

1 kilogram meter per square second (kg*m/s²) is equal to 1 Newton (N)
Our answer of 2 kg*m/s² is equal to 2 N

$F= 2 \ N$

The force is <u>2 Newtons.</u>

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Answer:

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Explanation:

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3 years ago

An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

REY [17]

Answer:

a) The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor ( $\alpha$ ), measured in radians per square second, as a function of time ( $t$ ), measured in seconds, is determined by the following formula:

$\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right]$ (1)

The function for the angular velocity of the electric motor ( $\omega$ ), measured in radians per second, is found by integration:

$\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right]$ (2)

Where $\omega_{o}$ is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration ( $a_{t}$ ), measured in meters per square second, is defined by the following formula:

$a_{t} = R\cdot \alpha$ (3)

Where $R$ is the radius of the electric motor, measured in meters.

If we know that $R = 7.165\times 10^{-2}\,m$ , $\alpha = 10\cdot t$ and $t = 1.5\,s$ , then the tangential component of the acceleration at the edge of the motor is: