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3 years ago

What change will always result in an increase in gravitational force between two objects

Physics

1 answer:

tekilochka [14]3 years ago

6 0

The gravitational force between two object depends on their masses and on their distance.

Since the formula is

$F = G\frac{m_1m_2}{d^2}$

If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.

So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.

So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

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How much potential energy does an 8 kg flower pot hanging 5 m above the ground have

Morgarella [4.7K]

Answer:

Explanation:

Relative to ground level it has

PE = mgh = 8(10)(5) = 400 J

8 0

2 years ago

Read 2 more answers

An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos

erica [24]

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

$\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm$

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

$\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm$

3 0

3 years ago

A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels

Kryger [21]

Answer:

T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

$T = \dfrac{v^2.m}{l}$

$v = \dfrac{l}{s}$

$v = \dfrac{7.2}{0.74}$

v = 9.73 m/s

now, calculation of tension

$T = \dfrac{9.73^2\times 0.46}{7.2}$

T = 6.0 N

The tension in the cord is equal to 6.0 N.

7 0

3 years ago

Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the

Westkost [7]

Answer:

They are able to balance torques due to gravity.

$F_1 L_1 = F_2L_2$

Explanation:

When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

$F_1 L_1 = F_2L_2$

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

since both friends are of different weight so they will balance themselves are different positions as per above equation

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3 years ago

How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for

kumpel [21]

As we know that electrostatic force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

$F_g = 1.98 \times 10^{20} N$

here we can assume that both will have equal charge of magnitude "q"

now we have

$1.98 \times 10^{20} = \frac{kq^2}{r^2}$

$1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}$

$1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2$

now we have

$q = 5.7 \times 10^{13} C$

6 0

3 years ago