Answer:
The Resultant Induced Emf in coil is 4∈.
Explanation:
Given that,
A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.
To find :-
find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).
So,
Emf induced in the coil represented by formula
∈ = 
...................(1)
Where:
. 
{ B is magnetic field }
{A is cross-sectional area}
. 
No. of turns in coil.
. 
Rate change of induced Emf.
Here,
Considering the case :-
& 
Putting these value in the equation (1) and finding the new emf induced (∈1)
∈1 =
∈1 =
∈1 =![4 [-N\times\frac{d\phi}{dt}]](https://tex.z-dn.net/?f=4%20%5B-N%5Ctimes%5Cfrac%7Bd%5Cphi%7D%7Bdt%7D%5D)
∈1 = 4∈ ...............{from Equation (1)}
Hence,
The Resultant Induced Emf in coil is 4∈.
Answer: a = 4 m/s²
Explanation:
a = Δv/t = (30 - 18) / 3 = 4 m/s²
Answer:
650.65 K or 377.5°C
Explanation:
Area = A = 10 m²
Thickness of wall = L = 2.5 cm = 2.5×10⁻² m
Inner surface temperature of wall = 
= 415°C = 688.15 K
Outer surface temperature of wall = 
Heat loss through the wall = 3 kW = 3×10³ W
Thermal conductivity of wall = k = 0.2 W/m K
Assumptions made here as follows
- There is not heat generation in the wall itself
- The heat conduction is one dimensional
- Heat flow follows steady state
- The material has same properties in all directions i.e., it is homogeneous.
Considering the above assumptions we use the following formula
∴ The temperature of the outer surface of the wall is 650.65 K or 377.5°C
Answer:
A) How much work does the Pepsi do on the bullet = 0.0625J
B) At what velocity does the Pepsi hit the floor = 7.67m/s
Explanation:
- Given mass of bullet = 5g
- initial velocity = 500 m/sec
From work done = Force X Distance
Workdone ; m(v - u )/t
A) The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi:
- but The work done by the container wall on the bullet is equal to the change in kinetic energy on either side of the wall: W = change in kinetic energy = 1/2 mv²
= 1/2 X 0.005 x 500² - 1/2 X 0.055 x 5²
= 624.94J
- Therefore ; The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi = W = change in KE = 0.0625 - 0 = 0.0625J
B) At what velocity does the Pepsi hit the floor?
- From conservation of energy principle; PE = KE
New sound intensity level after the move is
10 log (1/3²)
= -20 log (3)
= -20 (0.4771) = 9.54 dB LOWER than before the move.
