What TWO things affect the gravitational force acting on an object?

2 answers:

5 0

<u><em>mass and distance</em></u> (between two objects) are those TWO things that affect the gravitational force acting on an object

ohaa [14]2 years ago

3 0

The strength of the gravitational forces between two masses depends on

-- the product of the masses,

-- the distance between their centers of mass.

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A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.

Crazy boy [7]

A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N

b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds

6 0

3 years ago

What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass

Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

$F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz$

The frequency of the simple pendulum is given as;

$F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m$