Answer:
The terminal velocity of the diver is 115 m/s = 414 km/hr
Explanation:
At terminal velocity,
Fnet = mg - Fd = 0
Drag force, Fd = cρAv²/2
mg = cρAv²/2
Terminal Velocity of a body falling through a fluid as in a diver falling through air is given by
v = √(2mg/ρcA)
where m = mass of body falling through fluid = 80 kg
g = acceleration due to gravity = 9.8 m/s²
ρ = density fluid, density of air, as obtained from literature = 1.21 kg/m³
c = coefficient of drag friction of diver falling through air, as obtained from literature = 0.7
A = the area of the diver facing the fluid = 0.14 m²
v = √(2mg/ρcA) = √((2 × 80 × 9.8)/(1.21 × 0.7 × 0.14)) = 115 m/s = 115 × (3600/1000) km/hr = 414 km/hr
Answer:
B.The force of friction between the block and surface will decrease.
Explanation:
The force of friction is given by
where 
is the coefficient of friction and 
is the normal force.
When the student pulls on the block with force 
at an angle 
, the normal force on the block becomes
and hence the frictional force becomes
.
Now, as we increase , 
increases which as a result decreases the normal force 
, which also means the frictional force decreases; Hence choice B stands true.
<em>P.S: Choice D is tempting but incorrect since the weight </em>
<em> is independent of the external forces on the block. </em>
I believe if it were heavier with more mass, then the sun would pull it in and there would be no mercury. It might also be hotter.
Answer:
Work done is 0.
Explanation:
Given that,
The circumference of an orbit for a toy on a string is 18 m, r = 18 m
Centripetal force, F = 12 N
In the circular path, the centripetal force is always perpendicular to the motion of the object. Thus it makes an angle of 90 degrees with the force and displacement. Hence, we can say that the centripetal force does not do any work on the toy when it follows its orbit for one cycle.
