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3 years ago

Gravitational pull between 2 objects is determined by which of the following?

Physics

2 answers:

sp2606 [1]3 years ago

8 0

Answer:

Both distance between and mass of the 2 objects

Explanation:

The gravitational pull between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1 is the mass of the first object

m2 is the mass of the second object

r is the distance between the two objects

From the formula, we see that the gravitational pull depends both on the distance between the two objects (r) and on their masses (m1, m2), so the correct answer is

Both distance between and mass of the 2 objects

vampirchik [111]3 years ago

3 0

Both distance between and mass of the 2 objects.

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A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

IrinaK [193]

Answer : The specific heat capacity of the alloy $1.422J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$C_1$ = specific heat of alloy = ?

$C_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of alloy = 21.6 g

$m_2$ = mass of water = 50.0 g

$T_f$ = final temperature of system = $31.10^oC$

$T_1$ = initial temperature of alloy = $93.00^oC$

$T_2$ = initial temperature of water = $22.00^oC$

Now put all the given values in the above formula, we get

$21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC$

$c_1=1.422J/g^oC$

Therefore, the specific heat capacity of the alloy $1.422J/g^oC$

6 0

2 years ago

A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre

pantera1 [17]

Answer:

Part a)

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

$T = \frac{m}{L}xg + Mg$

so the speed of the wave at that position is given as

$v = \sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{L}$

now we have

$v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}$

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

time taken by the wave to reach the top is given as

$t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}$

$t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})$

$t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})$

$t = 12 s$

4 0

2 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

2 years ago

Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uni

nexus9112 [7]

Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

time of flight t = 2·V·sinθ/g

= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N

3 0

2 years ago

Read 2 more answers

What type of energy does the box have after it is done being pulled?

Ksivusya [100]

Assuming that the box is moving when it is being pulled, Work is done on the box.

So work is the Force times the distance

W=Fd

But what is work actually ? When something moves due to force over some change in distance, it have energy.

But where does this energy come from ? Does it magically appear ? The energy comes from the applied force onto the box.

So the energy have been transferred. And it’s like that throughout the universe

Now to save time, I’ll just tell you the answer: kinetic energy

:)

3 0

3 years ago