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Answer:
M
Explanation:
To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after
As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved
⇒
-
is the moment of interia of earth before impact -
is the angular velocity of earth about an axis passing through the center of earth before impact
is moment of interia of earth and asteroid system
is the angular velocity of earth and asteroid system about the same axis
let 
since 

⇒ if time period is to increase by 25%, which is
times, the angular velocity decreases 25% which is
times
therefore

(moment of inertia of solid sphere)
where M is mass of earth
R is radius of earth

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)
where
is mass of asteroid
⇒ 

=
+ 

⇒

Answer:
92.81 psia.
Explanation:
The density of water by multiplying its specific gravity by the density of sea water.
SG = density of sea water/density of water
ρ = SG x ρw
1 kg/m3 = 62.4 lbm/ft^3
= 1.03 * 62.4
= 64.27lbm/ft^3.
The absolute pressure at 175 ft below sea level as this is the location of the submarine.
P = Patm +ρgh
= 14.7 + 64.27 * 32.2 * 175
Converting to pound force square inch,
= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )
= 14.7 + 78.11 psia
= 92.81 psia.
Hi my friend, since momentum is always conserved without external forces, the momentum after the collosion will still be 0.06 kg*m/s. Hope it helps☺