Answer:
W = 9.6 N
Explanation:
Given that,
Area on 1 foot, A = 0.6 m²
Pressure, P = 16 Pa
The pressure is given by force acting per unit area. So,

So, the required weight is 9.6 N.
Answer:
i. The radius 'r' of the electron's path is 4.23 ×
m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r = 
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 ×
T, v = 121 m/s, Θ =
(since it enters perpendicularly to the field), q = e = 1.6 ×
C and m = 9.11 ×
Kg.
Thus,
r =
÷ sinΘ
But, sinΘ = sin
= 1.
So that;
r = 
= (9.11 ×
× 121) ÷ (1.6 ×
× 1.63 ×
)
= 1.10231 ×
÷ 2.608 × 
= 4.2266 ×
= 4.23 ×
m
The radius 'r' of the electron's path is 4.23 ×
m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f = 
= (1.6 ×
× 1.63 ×
) ÷ (2 ×
× 9.11 ×
)
= 2.608 ×
÷ 5.7263 × 
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.
Answer:
The tension on an object is equal to the mass of the object x gravitational force plus/minus the mass x acceleration. T = mg + ma.
Explanation:
Answer:
If the density of the object is high its molecular arrangement is compact while if the density is lows its molecular arrangement isnt that compact
Complete Question:
A coin is dropped off of a building landing on its side. It hits with a pressure of 400 N/m². It hits with a force of 0.1N. Calculate the area of the coin?
Answer:
Area = 0.00025 m²
Explanation:
Given the following data;
Pressure = 400N/m²
Force = 0.1N
To find the area of the coin;
Pressure = Force/area
Area = Force/pressure
Substituting into the equation, we have;
Area = 0.1/400
Area = 0.00025 m²