Answer:
v=0.60 m/s
Explanation:
Given that
m ₁= 390 kg ,u ₁= 0.5 m/s
m₂ = 250 kg ,u₂ = 0.76 m/s
As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.
Pi = Pf
m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v
Now putting the values in the above equation
390 x 0.5 + 250 x 0.76 = (390 + 250 ) v
v=0.60 m/s
Therefore the velocity of the system will be 0.6 m/s.
Explanation:
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Answer:
Explanation:
Given that,
Assume number of turn is
N= 1
Radius of coil is.
r = 5cm = 0.05m
Then, Area of the surface is given as
A = πr² = π × 0.05²
A = 7.85 × 10^-3 m²
Resistance of
R = 0.20 Ω
The magnetic field is a function of time
B = 0.50exp(-20t) T
Magnitude of induce current at
t = 2s
We need to find the induced emf
This induced voltage, ε can be quantified by:
ε = −NdΦ/dt
Φ = BAcosθ, but θ = 90°, they are perpendicular
So, Φ = BA
ε = −NdΦ/dt = −N d(BA) / dt
A is a constant
ε = −NA dB/dt
Then, B = 0.50exp(-20t)
So, dB/dt = 0.5 × -20 exp(-20t)
dB/dt = -10exp(-20t)
So,
ε = −NA dB/dt
ε = −NA × -10exp(-20t)
ε = 10 × NA exp(-20t)
Now from ohms law, ε = iR
So, I = ε / R
I = 10 × NA exp(-20t) / R
Substituting the values given
I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2
I = 1.67 × 10^-18 A
