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2 years ago

7

What will be the gravitational force between two heavenly bodies if the masses of both are tripled keeping the distance between

them constant

2 answers:

Leona [35]2 years ago

8 0

If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

Kruka [31]2 years ago

5 0

Answer:

If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

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N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$

B) the speed of the sphere with mass 107.0 kg is

$v_2=\frac{m_1v_1}{m_2}$

$=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s$

C) the magnitude of the relative velocity with which one sphere is

$v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s$

D) the distance of the centre is proportional to the acceleration

$\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962$

Thus,

$x_1=3.962x_2$

and

$x_2=0.252x_1$

When the sphere make contact with eachother

Therefore,

$x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r$

And

$x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r$

The point of contact of the sphere is

$32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m$

3 0

3 years ago

Which of the following equations describes photosynthesis?

BigorU [14]

Answer:

The chemical equation for photosynthesis is 6CO2+6H2O→C6H12O6+6O2.

Explanation:

4 0

2 years ago

A girl pulls on a 10-kg wagon with a constant horizontal force of 30 n. if there are no other horizontal forces, what is the wag

olga2289 [7]

Force = mass * acceleration
F = ma

Given m = 10 kg, F = 30 N;

F = ma
30 = 10a

Solving for a:
a = 3 m/s^2

The acceleration is 3 meters per second squared.

8 0

3 years ago

S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to

Gre4nikov [31]

By calculation, the diameter of the wire is 2.8 * 10^-3 m.

<h3>How do we obtain the length?</h3>

The following data are given in the question;

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

Area of the wire = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

Learn more about resistivity:brainly.com/question/14547003

#SPJ4

Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

6 0

1 year ago

A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

$\Omega = \frac{Mgd}{I\omega}$

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

$\omega$ = Angular velocity

Replacing the value for moment of inertia

$\Omega= \frac{MgR}{MR^2 \omega}$

$\Omega = \frac{g}{R\omega}$

The value for our angular velocity is not in SI, then

$\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})$

$\omega = 104.7rad/s$

Replacing our values we have that

$\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}$

$\Omega = 1.17rad/s$

The precession frequency is

$\Omega = \frac{2\pi rad}{T}$

$T = \frac{2\pi rad}{\Omega}$

$T = \frac{2\pi}{1.17}$

$T = 5.4 s$

Therefore the precession period is 5.4s

7 0

2 years ago