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2 years ago

7

What is the gravitational field strength at a distance of 60.0 km above the surface of the earth

1 answer:

zloy xaker [14]2 years ago

6 0

Answer:

g=9.64m/s^2.

Explanation:

Gravitational field strength (in other words, gravitational acceleration) is given as follows:g=GMR2g=R2GMwhere G=6.674×10−11m3kg⋅s2G=6.674×10−11kg⋅s2m3 is the gravitational constant, M=5.972×1024kgM=5.972×1024kg is the mass of the Earth, and R=6.371×106m+0.06×106m=6.431×106mR=6.371×106m+0.06×106m=6.431×106m is the distance from the center of the Earth to the required point above the surface (radius plus 60 km).

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gregori [183]

Answer:

-0.01 mm

Explanation:

We are given that

The value of one division of vernier scale =0.5 mm

The value of one main scale division=0.49 mm

We have to find the value of least count of the instrument in mm.

We know that

Leas count of vernier caliper=1 main scale division-1 vernier scale division

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Q3. A man walks with a speed of 8 m/sec in 40 sec. How much distance was covered?

chubhunter [2.5K]

Answer:

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A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900

Whitepunk [10]

It would be 17 m/s

If we use

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3 years ago

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A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between

Sunny_sXe [5.5K]

Answer : $4.483\times 10^{15}\ m/s^2$ .

Explanation:

It is given that,

Electric field strength, $E=2.55\times 10^{4}\ N/C$

We know that,

Charge of electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

From the definition of electric field, $F=qE$ ...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

$ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}$

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or

$a=4.483\times 10^{15}\ m/s^2$

So, the horizontal component of acceleration of an electron is $4.483\times 10^{15}\ m/s^2$ .

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3 years ago