<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.
The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)
The capacitance of a series combination is
1 / (1/A + 1/B + 1/C + 1/D + .....) .
If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is
(product of the 2 individuals) / (sum of the individuals) .
In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is
(1000 x 1) / (1000 + 1)
= (1000) / (1001)
= 0.999 000 999 . . .
which is smaller than the smaller individual.
It'll always be that way. </span>
Answer:
T=26.03 N
Explanation:
Given that
Distance between poles = 12 m
Mass of block m= 4 kg
Sag distance = 5 m
Lets take tension in the clothesline is T.
The component of tension in vertical direction will be T cosθ.
By force balancing
2 T cosθ = 40
here 
θ=39.80°
2 T cos39.8 = 40
T=26.03 N
Answer:
12.5 m/s
Explanation:
The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

Where, taking downward as positive direction, we have:
s = 8 m is the displacement of the hammer
u = 0 is the initial velocity (it is dropped from rest)
v is the final velocity
is the acceleration of gravity
Solving the equation for v, we find the final velocity:

So, the final speed is 12.5 m/s.
Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation

where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

hence, the change in Er is about 1.52J times the initial rotational energy
Answer: d) the presence of solidified lava flows on the Moon
Explanation:
A geological activity means an occurrence of event such as volcanic eruption, earthquake, sedimentation, erosion etc. The revolution of the Moon around the Earth
, the axial tilt of the Moon or the phases of the Moon are not surface features. hence, these events cannot provide the evidence of geological activity in the past of Moon.
The surface features of moon such as Mares, Craters, mountains, Rays and rills are the proof of some geological activity on the Moon. Mares are the dark patches on the moon's surface formed of solidified lava. Due to negligible atmosphere on the moon, the meteors strike its surface and cause craters to form. Thus, the correct answer is d.