Answer:
126000 J
Explanation:
Applying,
Q = cm(t₂-t₁).................. Equation 1
Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.
From the question,
Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C
Constant: c = 4200 J/kg.°C
Substitute these value into equation 1
Q = 2×4200(40-25)
Q = 2×4200×15
Q = 126000 J
For electrical devices . . .
Power dissipated = (voltage) x (current) =
(12 V) x (3.0 A) = 36 watts .
1 watt means 1 joule per second
(36 joule/sec) x (60 sec/min) x (10 min) = 21,600 joules
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.
That is true Step by step:
6.022*10^23 atoms/mole of reactant
(this is chemistry not physics)