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liq [111]
3 years ago
5

A roller coaster starts at the top of a straight track that is inclined at 30degrees with the horizontal. This causes it to acce

lerate at a rate of 4.9m/s^2(1/2g).
a. What is the roller coaster's speed after 3 seconds?
b. How far does it travel during that time?

Physics
2 answers:
Dahasolnce [82]3 years ago
8 0
Hope this helps you.

wlad13 [49]3 years ago
6 0

Answer:

Part a)

v_f = 14.7 m/s

Part b)

d = 22.05 m

Explanation:

Part a)

Initial speed of the roller coaster

v_i = 0

acceleration of the roller coaster is given as

a = 4.9 m/s^2

now for the speed after 3 s is given as

v_f = v_i + at

v_f = 0 + 4.9 \times 3

v_f = 14.7 m/s

Part b)

Distance traveled by the roller coaster

d = (\frac{v_f + v_i}{2})t

d = (\frac{14.7 + 0}{2})(3)

d = 22.05 m

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slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu
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Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

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Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

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3 years ago
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