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Veronika [31]
3 years ago
9

A space station has a mass M and orbits Earth in a circular orbit at a height above Earth’s surface. An astronaut in the space s

tation appears weightless because the astronaut seems to float. Which of the following claims is true about the force exerted on the astronaut by Earth ?
A) there is no force exerted on the astronaut by Earth because the astronaut is 400km above Earth’s surface

B) the force exerted on the astronaut by Earth is less than the force exerted by the astronaut because the astronaut is 400km above Earths surface

C) the force exerted on the astronaut by Earth is equal to the gravitational force of the space station that is exerted on the astronaut; the two equal forces balance to cause the astronaut to float.

D) the force exerted on the astronaut by Earth is equal to the force exerted on Earth by the astronaut

Please help me with this question if you can ! It would be very much appreciated . Thank you
Physics
1 answer:
Svetllana [295]3 years ago
8 0

Answer:

The correct option is;

D) The force exerted on the astronaut by Earth is equal to the force exerted on Earth by the astronaut

Explanation:

According to Newton's third law of motion, in nature, for every action, there is an equal and opposite reaction, such that if a first object exerts a certain amount of force on a second object, the second object will exert a force of equal magnitude and opposite direction to that exerted by the first object

Therefore, the gravitational force exerted by Earth on the astronaut, is equal to the force exerted by the astronaut on Earth.

You might be interested in
an auditorium measures 30.0 m ✕ 15.0 m ✕ 5.0 m. the density of air is 1.20 kg/m3. (a) what is the volume of the room in cubic fe
Fynjy0 [20]

dimension = 30.0 m ✕ 15.0 m ✕ 5.0 m.

density = 1.20 kg/m3

(a)volume = lenght * breadth * height

      = 30 * 15 * 5

      = 2250 metre cube = 2.25 cubic meter

(b)   mass of air = density * volume

        mass of air = 1.2 * 2250

mass of air = 2700kg

weight  = mass * 9.8

             = 2700 * 9.8

             = 26,460 N

  • The definition of Density is the amount of matter in a given space, or volume
  • Density = mass/volume
  • units for density kg/m^3
  • Density of water 1g/ml
  • Salt water is denser that is why  don't sink as easily.

To know more about density  visit : brainly.com/question/15164682

#SPJ4

5 0
1 year ago
A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
Mrac [35]

Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
2 years ago
Each second, 1250 m3 of water passes over a waterfall 150 m high. Three-fourths of the kinetic energy gained by the water in fal
mote1985 [20]

Answer:

The generator produces electrical energy at a rate of 1378125000 J per second.

Explanation:

volume of water falling each second is 1250 m^{3}

height through which it falls, h is 150 m

mass of 1 m^{3} of water is 1000 kg

⇒mass of 1250 m^{3} of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8 \frac{m}{sec^{2} }

in falling through 150 m in each second, by Work-Energy Theorem:

Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it

⇒KE = mgh

        = 1250000×9.8×150 J

        = 1837500000 J

Electrical Energy = \frac{3}{4}(KE)

                            = \frac{3}{4}×1837500000

                            = <u>1378125000 J per second</u>

8 0
2 years ago
Which of these is a type of matter that rarely interacts with other matter and is created in some nuclear fusion reactions withi
marysya [2.9K]

Answer:

the answer is helium you already know that because im in college on my way to the military next week bless me

Explanation:

7 0
3 years ago
At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon
ArbitrLikvidat [17]

Answer:

The  pressure at point 2 is P_2  = 254.01 kPa

Explanation:

From the question we are told that

   The speed at point 1  is  v_1  =  3.57 \ m/s

   The  gauge pressure at point 1  is  P_1  =  68.7kPa =  68.7*10^{3}\  Pa

    The density of water is  \rho  = 1000 \ kg/m^3

Let the  height at point 1 be  h_1 then the height at point two will be

      h_2  =  h_1  -  18.5

Let the  diameter at point 1 be  d_1 then the diameter at point two will be

      d_2  =  2 * d_1

Now the continuity equation is mathematically represented as  

         A_1 v_1  =  A_2 v_2

Here A_1 , A_2  are the area at point 1 and 2

    Now given that the are is directly proportional to the square of the diameter [i.e A=  \frac{\pi d^2}{4}]

   which can represent as

             A \ \  \alpha \ \  d^2

=>         A = c   d^2

where c is a constant

  so      \frac{A_1}{d_1^2}  =  \frac{A_2}{d_2^2}

=>          \frac{A_1}{d_1^2}  =  \frac{A_2}{4d_1^2}

=>        A_2  =  4 A_1

Now from the continuity equation

        A_1  v_1  =  4 A_1 v_2

=>     v_2  =  \frac{v_1}{4}

=>     v_2  =  \frac{3.57}{4}

       v_2  =  0.893 \  m/s

Generally the Bernoulli equation is mathematically represented as

       P_1 + \frac{1}{2}  \rho v_1^2 +  \rho *  g * h_1  =  P_2 + \frac{1}{2}  \rho v_2^2 +  \rho *  g * h_2

So  

         P_2  =  \rho  * g  (h_1 -h_2 )+P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )  

=>    P_2  =  \rho  * g  (h_1 -(h_1 -18.3)  + P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )

substituting values

        P_2  =  1000  * 9.8  (18.3) )+ 68.7*10^{3}  +  \frac{1}{2}  *  1000 ((3.57)^2 -0.893 ^2 )

       P_2  = 254.01 kPa

 

8 0
2 years ago
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