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3 years ago

If two twins (54 kg each) were 0.02 m apart, what is the force of gravity between them?

Physics

1 answer:

Lelechka [254]3 years ago

3 0

Answer:

Force, $F=4.86\times 10^{-4}\ N$

Explanation:

We have,

Masses of two twins are 54 kg each

They are placed at a distance of 0.02 m

It is required to find the force of gravity between them. The formula used to find the gravitational force between masses is given by :

$F=G\dfrac{m_1m_2}{r^2}$

plugging all the known values:

$F=6.67\times 10^{-11}\times \dfrac{54^2}{(0.02)^2}\\\\F=4.86\times 10^{-4}\ N$

So, the force of gravity between them is $4.86\times 10^{-4}\ N$ .

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What do neon,oxygen and nitrogen have in common?

julsineya [31]

Their Period number is common means their "Principal Quantum Numbers" are same

Hope this helps!

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3 years ago

A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i

Effectus [21]

Answer:

s = 1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}

• s = $\sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}$

s = $\sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}$

s = 1.7 m

7 0

3 years ago

A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the ground. How far, in meters, from the deck doe

Vesna [10]

Answer:

2.605m

Explanation:

Using the formula for calculating Range (distance travelled in horizontal direction)

Range R = U√2H/g

U is the speed = 4.8m/s

H is the maximum height = ?

g is the acc due to gravity = 9.8m/s²

R = 3.5m

Substitute into the formula and get H

3.5 = 4.8√2H/9.8

3.5/4.8 = √2H/9.8

0.7292 = √2H/9.8

square both sides

0.7292² = 2H/9.8

2H = 0.7292² * 9.8

2H = 5.21

H = 5.21/2

H = 2.605m

Hence the height of the ball from the ground is 2.605m

7 0

3 years ago

What is the weight of a 42 kg object if the object was on the moon?

sveticcg [70]

(1.6 m/s²)(42 Kg)= 80 N

4 0

4 years ago

A ball is bouncing to a height that is 1/3 of the previous height after every bounce. What kind of sequence is this? How could y

tamaranim1 [39]

Answer and Explanation:

The ball is bouncing to a height of 1/3 of its previous height this is a type of geometric sequence the total distance can be found by the sum of geometric sequence

For example let the initial height is 243 fit

After one bounce it will reach 243/3 =81 feet

After second bounce 81/3=27 feet

After third bounce 27/3 =9 feet

After fourth bounce 9/3 =3 feet

So a sequence is formed that is 243,81,27,9,3..........

Here $r=\frac{3}{9}=\frac{1}{3}$

Sum of infinite GP = $\frac{a}{1-r}$

From this formula we can find the total distance traveled by the ball

3 0

3 years ago