Answer:
Explanation:
(a)Gravitational Force is W=mg
Frictional force is f
Normal force is N
(b)Since the block is in Equilibrium therefore from diagram we can see that



(c)Magnitude of Normal Force

(d)friction is also given by

and 
thus 
Answer:
e. TA>T>Tc
Explanation:
a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:

if we did an energy balance of the system, we would get that>
QA+QB+QC=0
For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.
We don't know either of them, so we cannot determine if this statement is true.
b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.
c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.
d) is not complete
e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.
m = mass of the partner which the cheerleader lifts = 59.6 kg
h = height to which the partner is lifted by the cheerleader = 0.749 m
g = acceleration due to gravity = 9.8 m/s²
work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.
W = work done by the cheerleader in lifting the partner
PE = potential energy gained
so W = PE
potential energy is given as
PE = mgh
hence
W = mgh
inserting the values in the above formula
W = 59.6 x 9.8 x 0.749
W = 437.5 J
this is the work done in lifting the partner once.
the cheerleader does this 30 times , hence the total work done is given as
W' = 30 W
W' = 30 x 437.5
W' = 13125 J
Answer:
option D
Explanation:
given,
wavelength = 600 nm
width of separation = 0.02 mm
L = 5 m
for mth order maxima

for (m+1)th order maxima

now,
and

hence,





hence, the correct answer is option D