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lesya [120]
3 years ago
9

a truck starts to move from rest if it given the acceleration of 3 metre per second square in 5 second calculate the final veloc

ity and distance covered by truck in full process​
Physics
1 answer:
frosja888 [35]3 years ago
7 0
Hope this helps. Plzz let me know if you have any further doubts :)

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6.71081mph(miles per hour) or 3m/s (meters per second)
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The ball drop 2kms in the air
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Velocity is:
sleet_krkn [62]

Answer:

d

Explanation:

Solution:-

- The Quantity of theory of money states:

                      M * V = P * Y

Where,

           M = Money supply

           V = Velocity of money exchange

           P = The price level

           Y = Real GDP

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A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth
Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

W=mg

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

g=\frac{GM}{r^2}

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center


The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

800 N= W=mg=m \frac{GM}{R^2} (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

200 N = W' = mg' = m \frac{GM}{(R+h)^2} (2)

If we divide eq.(1) by eq.(2), we get

\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}

4=\frac{(R+h)^2}{R^2}

By solving the equation, we find:

4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0

which has two solutions:

h=-3R --> negative solution, we can ignore it

h=R --> this is our solution

Since the Earth's radius is R=6.4\cdot 10^6 m, the person should be at h=R=6.4\cdot 10^6 m=6400 km above Earth's surface.

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4 years ago
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