Answer:
No. 67
Peter Street
12th Road
Chennai
24th June 201_
Dear Amrish
I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.
It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?
I am anxiously waiting for your reply.
Yours affectionately
your name
Answer:
3.75 × 10⁻⁸ N
Explanation:
Given:
Intensity of the electromagnetic wave, I = 150 W/m²
Sides of the board = 25 cm (= 0.25 m) and 30 cm (= 0.30 m)
therefore,
the area of the rectangular box, A = 0.25 × 0.30 = 0.075 m²
Now,
force exerted on the card by the radiation, F =
here,
C is the speed of the light = 3 × 10⁸ m/s
on substituting the respective values, we get
F =
or
F = 3.75 × 10⁻⁸ N
Explanation:
Mass of the astronaut, m₁ = 170 kg
Speed of astronaut, v₁ = 2.25 m/s
mass of space capsule, m₂ = 2600 kg
Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :
initial momentum = final momentum
Since, initial momentum is zero. So,



So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.
Answer:
Energy stored in the capacitor is
Explanation:
It is given that,
Charge, 
Potential difference, V = 36 V
We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

U = 0.000027 J

So, the potential energy is stored in the capacitor is
. Hence, this is the required solution.
Answer:

Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e (
)
with this expression we can find the closest approach distance (r)