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3 years ago

A satellite is in orbit 3.110106 m from the center of Earth. The mass of Earth is 5.98011024 kg Calculate the orbital

Physics

1 answer:

TEA [102]3 years ago

3 0

Answer:

9194.4278 seconds

Explanation:

Orbital period is given by :

T² = (4π²r³) /GM

G = Gravitational constant = 6.67 * 10^-11

M = 5.9801 * 10^24 kg

r = (3.110 * 10^6 m + 6378000 m) =3110000 + 6378000 = 9488000 m

T² = (4π² * 9488000^3) / (6.67* 10^-11 * 5.9801*10^24)

T² = 3.37197 * 10^22 / 398872670000000

T² = 84537504.161415

T = sqrt(84537504.161415)

T = 9194.4278 seconds

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(b) if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

The force on an object is determined by applying Newton's second law of motion;

F = ma

$\frac{F_1}{m_1a_1} = \frac{F_2}{m_2a_2}$

(a)

when the force is halved, F₂ = 0.5F₁,

mass is doubled, m₂ = 2m₁

$\frac{F_1}{m_1a_1} = \frac{0.5F_1}{2m_1a_2} \\\\2m_1a_2F_1 = 0.5F_1 m_1a_1\\\\2a_2 = 0.5a_1\\\\a_2 = \frac{0.5a_1}{2} = \frac{a_1}{2 \times 2} = \frac{a_1}{4} \\\\a_2 = \frac{1}{4} (a_1)$

Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)

when the force is doubled, F₂ = 2F₁,

mass is halved, m₂ = 0.5m₁

$\frac{F_1}{m_1 a_1} = \frac{2F_1}{0.5m_1 a_2} \\\\0.5m_1a_2 F_1 = 2F_1m_1a_1\\\\0.5a_2 = 2a_1\\\\a_2 = \frac{2a_1}{0.5} \\\\a_2 = 4(a_1)$

Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

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2 years ago

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3 years ago

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2 years ago

An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a uniform volume charge density p. Dete

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Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;

c) r>b E=ρ b (b-a)/r*εo

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As we know,

∫E.dr= Qinside/εo

For r<a --->Qinside=0 then E=0

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2 years ago

A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.

Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

$v^{2}=u^{2}+2as$

$0^{2}=35^{2}- 2 \times 9.8 \times h$

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

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The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

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3 years ago