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3 years ago

I really need some with help with this anyone who can help with acceleration and velocity pleasee and thank you

Physics

1 answer:

kykrilka [37]3 years ago

3 0

Answer:

velocity equals distance divided by time and the unit you use after you get the answer is m/s

Explanation:

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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 :

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

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

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

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

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

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

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

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

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

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

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

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

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

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

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

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

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

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

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3 years ago

A tennis ball is dropped from a roof. If it takes 38.9 seconds to reach the ground, how fast is the ball moving just before it h

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Answer:

The velocity of the ball before it hits the ground is 381.2 m/s

Explanation:

Given;

time taken to reach the ground, t = 38.9 s

The height of fall is given by;

h = ¹/₂gt²

h = ¹/₂(9.8)(38.9)²

h = 7414.73 m

The velocity of the ball before it hits the ground is given as;

v² = u² + 2gh

where;

u is the initial velocity of the on the root = 0

v is the final velocity of the ball before it hits the ground

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 7414.73 )

v = 381.2 m/s

Therefore, the velocity of the ball before it hits the ground is 381.2 m/s

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