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UNO [17]
4 years ago
6

If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gr

avity at the surface would becomeA) 1/2 of what it now is.B) 2 times what it now is.C) 1/4 of what it now is.D) the same as it now is.E) 4 times what it now is.
Physics
1 answer:
Lerok [7]4 years ago
5 0

Answer:

B) 2 times what it now is

Explanation:

The acceleration due to gravity at the surface of the Earth is given by

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

where

G is the gravitational constant

M is the mass of the Earth

R is the Earth's radius

In this problem, the mass of the Earth is doubled:

M' = 2M

while the radius remains the same:

R' = R

so the new acceleration due to gravity would be

g'=\frac{GM'}{R'^2}=\frac{G(2M)}{R^2}=2\frac{GM}{R^2}=2g

so, the acceleration due to gravity would become twice the current value.

Note also that the value of g does not depend on the mass of the objects involved.

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A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
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Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
4 years ago
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