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EastWind [94]
2 years ago
15

. The international space station has a mass of 4.2 x 105 kg and it orbits the Earth at an average altitude of 400 km above the

Earth's surface. The radius of the Earth is 6400 km, and the mass of the Earth is 6.0 x 1024 kg. Assume the only force acting on the space station is the Universal Gravitational Force, Fo, and assume that the orbit is perfectly circular. How long would it take for the space station to make one complete orbit around the Earth given these assumptions? A. 93 minutes B. 3.5 x 10 minutes C. 85 minutes D. 21 minutes E. 1.3 minutes
Physics
1 answer:
Tresset [83]2 years ago
5 0

Answer:

A. 93 min

Explanation:

M = mass of earth = 6.0 x 10²⁴ kg

R = Radius of earth = 6400 km = 6.4 x 10⁶ m

h = Altitude above earth = 400 km = 0.4 x 10⁶ m

Radius of orbit of the space station around earth is given as

r = R + h

r = 6400 + 400

r = 6800 km

r = 6.8 \times 10^{6} m

T = Time period of orbit of space station

Here we can use Kepler's third law which is given as

T^{2} = \frac{4\pi ^{2} r^{3}}{GM}

Inserting the above values

T^{2} = \frac{4(3.14)^{2} (6.8\times10^{6})^{3}}{(6.67\times10^{-11})(6.0\times10^{24})}

T^{2} = 3.1\times 10^{7}

T = 5566.53 sec

T = 5566.53 \left ( \frac{1min}{60sec} \right )= 93 min

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What acceleration results from exerting a 125N force on a 0.65kg ball?
labwork [276]
first you do your pyramid f is on top and ma is
on bottom were m=mass and a=acceleration
were in this case you do f÷m=force÷mass so again in this case 125÷0.65=192.3 to be more accurate 192.3076923077
5 0
3 years ago
The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c
Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

\Sigma F = F_{E}-W = 0 (Eq. 1)

Where:

F_{E} - Electrostatic force exerted on human, measured in Newton.

W - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

q\cdot E-m\cdot g = 0

q = \frac{m\cdot g}{E} (Eq. 2)

E - Electric field, measured in Newtons per Coloumb.

m - Mass, measured in kilograms.

g - Gravity acceleration, measured in meters per square second.

q - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that m = 60\,kg, g = 9.807\,\frac{m}{s^{2}} and E = -150\,\frac{N}{C}, the charge that a 60-kg human must have to overcome weight is:

q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }

q = -3.923\,C

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

F = \kappa \cdot \frac{q^{2}}{r^{2}} (Eq. 3)

Where:

\kappa - Electrostatic constant, measured in Newton-square meter per square Coulomb.

q - Electric charge, measured in Coulomb.

r - Distance between two people, measured in meters.

If we know that \kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q = -3.923\,C and r = 100\,m, then the force of repulsion between two people is:

F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]

F = 13.851\times 10^{6}\,N

The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

5 0
3 years ago
Please need help with this
Stella [2.4K]

Answer:

D. The objects particles move faster

4 0
3 years ago
Read 2 more answers
Pls help it's due today​
mario62 [17]

Answer:

Breh seriously. Ugh fine.

1.B

2.D

3.C

4.C

5.D,A and B

6.A,C and D

3 0
2 years ago
Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.
Strike441 [17]

Answer:

The potential difference between the ends of a wire is 60 volts.

Explanation:

It is given that,

Resistance, R = 5 ohms

Charge, q = 720 C

Time, t = 1 min = 60 s

We know that the charge flowing per unit charge is called current in the circuit. It is given by :

I = 12 A

Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :

V = IR

V = 60 Volts

So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.

8 0
2 years ago
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