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
The angular speed can be solve using the formula:
w = v / r
where w is the angular speed
v is the linear velocity
r is the radius of the object
w = ( 5 m / s ) / ( 5 cm ) ( 1 m / 100 cm )
w = 100 per second
Answer:
W = 222 N.
Explanation:
The qiestion says" If the acceleration of gravity on the surface of the planet Mercury is 3.7 m / s2, then what would be the weight of a person with mass 60 kg on its surface?
"
Mass of the person, m = 60 kg
The acceleration due to gravity on the surface of gravity is 3.7 m/s²
We need to find the weight of a person on the surface of Mercury.
Weight of an object is given by :
W = mg
So,
W = 60 kg × 3.7 m/s²
W = 222 N
Hence, the person will weigh 2222 N on the surface of Moon.
Explanation:
googlmeet join community cet-uije-bqs
<span>Answer:
initial I = (m/2)L²/3 + (m/2)L²
where L = ½ the length of the rod, and the vertical half can be treated as a point mass.
initial I = mL²(1/6 + 1/2) = 2mL²/3
final I = m(2L)²/3 = 4mL²/3
Since I has doubled and momentum is conserved, ω has halved.
ω = 3.9 rad/s.
Formulaically: 2mL²/3 * 7.8rad/s = 4mL²/3 * ω</span>
