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USPshnik [31]
3 years ago
11

A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.

2 km/s. The minimum speed needed to escape from the surface of the planet is 14.9 km/s, and G = 6.67 × 10-11 N · m2/kg2. The orbital period of the satellite is closest to
Physics
1 answer:
Ksju [112]3 years ago
3 0

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

v =  \sqrt{\frac{2GM}{R}}

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

v = 14.9km/h = 14900m/s

Through this equation we can find the mass of the Planet in function of the distance, therefore

M = \frac{v^2R}{2G}

M = \frac{14900^2R}{2(6.67*10^{-11})}

M = 16.64*10^{17}R

The orbital velocity is

v_o = \sqrt{\frac{GM}{R+h}}

9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}

11.1*10^7R = (R+15000*10^3)(9200)^2

2.64*10^7R = 12.69*10^{13}

R = 4.81*10^6m

The time period of revolution is,

T = \frac{2\pi(R+h)}{v_o}

T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}

T = 4307s

T = 72min = 1hour12min

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

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nirvana33 [79]

Answer:

a) F = 882.63\,N, b) \dot W= 4413.15\,W, c) \eta = 15.216\,\%.

Explanation:

a) Let assume that car travel on a horizontal surface. The equations of equilibrium of the car are:

\Sigma F_{x} = F - \mu_{r}\cdot N = 0

\Sigma F_{y} = N - m\cdot g = 0

After some algebraic handling, the following expression for the propulsion force is constructed:

F = \mu_{r}\cdot m \cdot g

F = (0.06)\cdot (1500\,kg)\cdot (9.807\,\frac{m}{s^{2}} )

F = 882.63\,N

b) The power require to move the car at a speed of 5 meters per second is:

\dot W = F\cdot v

\dot W = (882.63\,N)\cdot (5\,\frac{m}{s} )

\dot W= 4413.15\,W

c) The efficiency of the car is:

\eta = \frac{(882.63\,N)\cdot (15\,mi)\cdot (\frac{1609\,m}{1\,mi} )}{(1.4\times 10^{8}\,J)} \times 100\,\%

\eta = 15.216\,\%

3 0
3 years ago
Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

U=k\frac{q_1q_2}{r_{1,2}}                  (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}           (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J

The electric potential energy between the three charges is 80.91 J

7 0
3 years ago
A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60
miskamm [114]

Answer:

Answer:196 Joules

Explanation:

Hello

Note:  I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work  is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules

Answer:196 Joules

I hope it helps

5 0
3 years ago
Which bright object is in shadow
IRINA_888 [86]
My best guess would be sun because it is bright but is surrounded by shadows on all sides.
5 0
3 years ago
collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitu
anzhelika [568]

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

4 0
3 years ago
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