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3 years ago

What happens to the force of gravitational pull if we tripled the distance and one of the masses is doubled

2 answers:

5 0

Triple the distance ... The force drops to 1/9 of the original force.

Double one mass ... The original force doubles.

Do both ... The original force becomes 2/9 as strong, = about 22.2% of what it was originally.

kicyunya [14]3 years ago

4 0

This depends on the original mass of the object having its mass doubled and the the original distance before the distance was tripled.

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The gravity of Neptune is about 1.1 times the gravity of earth. How will the mass of an object on Neptune compare with its mass

Roman55 [17]

I think the gravity doesn't affect the mass of an object. Only it's weight can be compared

7 0

2 years ago

Read 2 more answers

a 100 kg person travels from sea level to an altitude of 5000 m. By how mans newtons does their weight change?

Bumek [7]

Answer:

Thus, the change in the weight of the person is 1.6N , option c is correct.

Explanation:

3 0

1 year ago

What is the relationship between these animals (Predator-Prey) (Parisite-Host) (mutualism) or (commensalism)

ohaa [14]

1.commensalism
2. pred-prey
3. parasite-host
4.commensalism

8 0

3 years ago

A jet plane lands with a velocity of +107 m/s and can accelerate at a maximum rate of -5.18 m/s2 as it comes to rest. From the i

Marrrta [24]

Answer:

20.7 s

Explanation:

The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:

v = a*t + v₀

Solve for t:

t = (v - v₀)/a

3 0

2 years ago

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

7 0

3 years ago