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3 years ago

Starting from rest, a rocket accelerates at 20 m/s². How far will it travel in the first 4

Physics

1 answer:

guajiro [1.7K]3 years ago

3 0

Answer:

80m

Explanation:

using

S=ut+1/2at².

u=0

t=4s

v=20 m/s².

s=0Xt+1/2x10x4².

S=5x16

=80m

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uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0

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3 years ago

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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

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 :

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

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

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

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

= 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

3 0

3 years ago