Answer:
194516 sheets
Explanation:
So the area of each sheet of paper is:
A = 0.216 * 0.279 = 0.060264 square meters
For the paper sheet to make the same effect as the atmospheric pressure P, then the gravity F from the paper sheet must be
F = AP = 0.060264 * 101325 = 6106 N
Let g = 9.81 m/s2, then the mass of paper needed to generate that gravity is
m = F/g = 6106 / 9.81 = 622.4 kg
If each sheet has a mass of 0.0032 kg, then the total number of sheets to have that much mass is
622.4 / 0.0032 = 194516 sheets
Answer:
C
Explanation:
The answer is C) the mass of an object
Answer: Third option
F = 250w
Explanation:
The impulse can be written as the product of force for the time interval in which it is applied.
You can also write impulse I as the change of the linear momentum of the ball
So:
We want to find the force applied to the ball. We know that
milliseconds = 0.03 seconds
The initial velocity 
is zero.
The final speed 
So
We must express the result of the force in terms of the weight of the ball.
We divide the expression between the acceleration of gravity
The answer is the third option
Answer:
After finding the electric potential VP at point P = Q/Чπϵ₀L ㏑(1+
)
Explanation:
I believe it is a part C question.
The derivative of V and P will be directly proportional to the differential dq and the inverse of Чπϵ₀δ........
Please find detailed solution in the attached picture as i believe that is the answer to the part C question you are seeking for.
Answer:
unknow e and f
Explanation:
In experiments with alpha particles that are obtained by the method of radioactive decay of atoms, some parameters are known
a) Known. The initial velocity is given by the energy of the particles entities by the atomic nuclei
b) Known. The particle charge always 2e, helium core
c) Known. It is set in the given experiment, in general it is selected as zero
d) Known. Placed by the experimenter
e) Unknown. The speed depends on the interactions with the system
f) Unknown. It depends on the interactions with the system, because the position depends on the interactions
g) Known. It is always the value of a helium nucleo
