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creativ13 [48]
2 years ago
5

If a car is moving at a constant velocity of 10 m/s, what is its acceleration?

Physics
1 answer:
marin [14]2 years ago
4 0

Answer:

If a car is moving at a <em><u>constant velocity of 10 m/s,</u></em> there will be no change in the velocity per time.

Acceleration is the rate of change of velocity

Acceleration=(Final velocity-Initial velocity)/Time

Acceleration=(10m/s-10m/s)/Time= <em><u>0 m/s²</u></em>

<h3>★ <u>0 m/s²</u> is the right answer. </h3>
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Weightlessness is experienced by an astronaut in space. This means that the astronaut's muscles have to be stronger to move his
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The answer is false. The speed of the astronaut cancels out the force of gravity, causing a 'stationary freefall'. While under these effects, it is not required for an astronaut to 'strengthen' his body.
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3 years ago
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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
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Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

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Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

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3 0
3 years ago
20. In a parallel RL circuit, 10 mA flows through the resistor and 4 mA flows through the inductor. What phase angle separates v
stellarik [79]

Answer:

Option D: 21.8 degrees

Explanation:

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Therefore, the phase angle that separated output voltage and total current can be obtained using the fact that tan(phase angle) = \frac{I_l}{I_R} = \frac{x}{y} \frac{4}{10}, therefore the angle is the arctangent of 4/10:

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