Answer:
129 J/Kg°C
Explanation:
Given :
Mass of gold, m = 1.2kg
Quantity of heat applied, Q = 3096 J
Temperature, t2 = 40°C
Temperature, t1 = 20°C
Change in temperature, dt = (40-20)°C = 20°C
Using the relation :
Q = mcdt
Where, C = specific heat capacity of gold
3096 = 1.2kg * C * 20°C
3096 J = 24kg°C * C
C = 3096 J / 24 kg°C
C = 129 J/Kg°C
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
Answer with Explanation:
We are given that
A=3i-3j m
B=i-4 j m
C=-2i+5j m
a.
Compare with the vector r=xi+yj
We get x=2 and y=-2
Magnitude=
units
By using the formula 
Direction:
By using the formula
Direction of D:
b.E=-A-B+C
units
Direction of E=
