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mezya [45]
1 year ago
5

a 1300-kilogram space vehicle travels at 4.8 meters per second along the level surface of mars. if the magnitude of the gravitat

ional field strength on the surface of mars is 3.9 newtons per kilogram, the magnitude of the normal force acting on the vehicle is (1) 320 n (2) 4400 n (3) 930 n (4) 5070 n
Physics
1 answer:
Helga [31]1 year ago
5 0

The magnitude of the normal force acting on the space vehicle will be 5070 newtons. So option (4) is correct.

Given that,

The mass of the space vehicle is 1300 kilograms

It travels at a speed of 4.8 meters per second along the level surface of mars.

The g' (gravitational field strength) on the surface of mars is 3.9 newtons per kilogram

We need to find the normal force acting on the space vehicle

We know, N = mg' (where N is normal force)

N = (1300 × 3.9) Newton

N = 5070 Newton

So, the magnitude of the normal force acting on the space vehicle that is traveling at 4.8 meters per second along the level surface of mars will be 5070 newtons.

The normal force is the force that surfaces exert to prevent solid objects from passing through each other. The normal force is one type of ground reaction force.

Learn more about the normal force here:

brainly.com/question/6922206

#SPJ4

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at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
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The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

where

m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

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