Answer:
100 Joule
Explanation:
Amount of heat in agiven body is given by Q = m•C•ΔT
where m is the mass of the body
c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.
ΔT is the change in the temperature of body
___________________________________________
coming back to problem
m = 5g
C = 2J/gC
since, it is given that temperature of body increases by 10 degrees, thus
ΔT = 10 degrees
Using the formula for heat as given
Q = m•C•ΔT
Q = 5* 2 * 10 Joule= 100 Joule
Thus, 100 joule heat must be added to a 5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.
Answer: B) The ability of steel to be shaped
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.