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KengaRu [80]
3 years ago
12

The free-fall acceleration on Mars is 3.7 m/s2.

Physics
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

(a) The length of the pendulum on Earth is 36.8cm

(b) The length of the pendulum on Mars is 13.5cm

(c) Mass suspended from the spring on Earth is 0.37kg

(d) Mass suspended from the spring on Mars is 0.36kg

Explanation:

Period = 1.2s, free fall acceleration on Earth = 9.8m/s^2, free fall acceleration on Mars = 3.7m/s^2

( a) Length of pendulum on Earth = [( period ÷ 2π)^2] × acceleration = (1.2 ÷ 2×3.142)^2 × 9.8 = 0.0365×9.8 = 0.358m = 35.8cm

(b) Length of the pendulum on Mars = (1.2÷2×3.142)^2 × 3.7 = 0.0365×3.7 = 0.135cm = 13.5m

(c) Mass suspended from the spring on Earth = (force constant×length in meter) ÷ acceleration = (10×0.358) ÷ 9.8 = 0.37kg

(d) Mass suspended from the spring on Mars = (10×0.135)÷3.7 = 0.36kg

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Can someone solve this problem and explain to me how you got it​
Zarrin [17]

1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

1)

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

r' = 5r

Therefore, the new force between the charges is:

F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

F=k\frac{q_1 q_2}{r^2}

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

- The charges are both changed by a factor of 8:

q_1' = 8q_1

q_2' = 8q_2

Substituting into the equation, we find the new force:

F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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