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4 years ago

What is the definition of permitted orbits ?

1 answer:

7 0

.The path of a celestial body or an artificial satellite as it revolves around another body due to their mutual gravitational <span>attraction.</span>

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A charged particle is accelerated in a uniform electric field. When its velocity is 2 m/s, its electric potential energy is 100

zavuch27 [327]

Answer:

particle's potential energy = 70J

Explanation:

From conservation of energy; K1 + Ue1 = K2 + Ue2

where K1 and K2 are the kinetic energies at two positions and Ue1 and Uue2 are the electrical potential energies at two positions.

k1 = 10J, Ue1 = 100J

K2 = 40J

substitute into K1 + Ue1 = K2 + Ue2

Ue2 = K1 + Ue1 - K2

= 10 +100 - 40

Ue2 = 70J

7 0

3 years ago

Atoms contain empty space true or false

olga nikolaevna [1]

The answer would be false

5 0

4 years ago

Young's Modulus of elasticity is a) Shear stress/Shear strain b) Tensile stress/Shear strain 9. c) Shear stress /Tensile strairn

slava [35]

Answer:

Option C is the correct answer.

Explanation:

Young's modulus is the ratio of tensile stress and tensile strain.

Bulk modulus is the ratio of pressure and volume strain.

Rigidity modulus is the ratio of shear stress and shear strain.

Here we are asked about Young's modulus which is the ratio of tensile stress and tensile strain.

Option C is the correct answer.

3 0

3 years ago

Read 2 more answers

There is a basketball with a diameter of 9in. and a softball with a diameter of 4in. Find how many times greater the volume of t

erica [24]

Answer: #.37 times greater

Explanation: Volume of a sphere is: V = $\frac{4}{3}$ ·π·r³

Since it is given the diameter of the basketball and the softball, their ratios are r₁ = 3in and r₂ = 2in, in which index 1 represents the basketball and index 2, the softball.

Calculating the volume of basketball:

V₁ = $\frac{4}{3}$ . 3.14 . 3³

V₁ = 113.04 in³

Calculating the volume of softball:

V₂ = $\frac{4}{3}$ . 3.14 . 2³

V₂ = 33.5 in³

Comparing one volume with the other:

$\frac{V1}{V2}$ = $\frac{113.04}{33.5}$ = 3.37

We have that V₁ = 3.37V₂

This means the volume of a basketball is 3.37 times greater than the volume of a softball.

4 0

3 years ago

Read 2 more answers

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

6 0

4 years ago