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2 years ago

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 constant velocity of 10 m/s, 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= 0 m/s²

<h3>★ 0 m/s² 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

Ber [7]

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 :

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

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

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

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 :

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

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

$= 3371 \times 153.86 \times 10^{-8}$

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Acceleration due to gravity at 4000 km depths is :

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

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

$= 153.84 \times 2371 \times 10^{-8}$

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= 3.64 m/s

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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:

In a parallel RL circuit, the current in the resistor R and that in the inductor L are separated among themselves 90 degrees as illustrated in the attached image. In the image the current in the resistor is represented in orange, that of the inductor in blue, and the total current (vector addition of the previous two) is represented in red, forming a certain angle (theta) with respect to the current in the resistor. The output voltage is the same as the input voltage as measured over the resistor R.

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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