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3 years ago

13

If you neglect air resistance, the horizontal component of a projectile A. Is always changing B. Remains constant throughout the

flight C. Increases, goes to zero, then increases D. Depends on who threw it

1 answer:

Shtirlitz [24]3 years ago

8 0

Answer:

B. Remains constant throughout the flight

Explanation:

If we completely neglect air resistance then the projectile will not have any horizontal resistances to deal with, therefore the horizontal component will remain constant throughout the flight. This would continue to be the case until it meets some form of resistance. Which assuming that everything else is normal would be the case since the force of gravity will push the projectile down (vertical component) until it hits the ground which in that instance would act as an opposing force to the horizontal component as well.

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A ball kept on 3rd floor of a building.

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A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant

atroni [7]

Answer:

The time constant is $\tau = 1.265 s$

Explanation:

From the question we are told that

the time take to charge is $t = 2.4 \ s$

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$V = V_o - e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant

$V_o$ is the potential of the capacitor when it is full

So the capacitor potential will be 100% when it is full thus $V_o =$ 100% = 1

and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

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3 years ago

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 :

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

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

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Acceleration due to gravity at 2000 km depths is :

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

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

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

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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