Answer:
Ein: 2.75*10^-3 N/C
Explanation:
The induced electric field can be calculated by using the following path integral:
Where:
dl: diferencial of circumference of the ring
circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m
ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)
The electric field is always parallel to the dl vector. Then you have:
Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:
hence, the induced electric field is 2.75*10^-3 N/C
Answer:
Hacksaw blade because it is metal.
Answer:
6250 J
Explanation:
Work is the distance times the component of the force along that distance.
The horizontal component of the force is:
25 N × cos(60°) = 12.5 N
So the work done is:
W = (500 m) (12.5 N)
W = 6250 J
Answer:
a
b
Explanation:
From the question we are told that
The spring constant is 
The mass of the block is 
Considering question a
Generally the weight of the block is mathematically represented as
=> 
=> 
Generally the force exerted on the spring is mathematically represented as
At equilibrium
=> 
=> 
=>
Considering question b
Generally the energy stored in the spring is mathematically represented as
Now the potential energy of the block before it is drooped is mathematically represented as
Generally from the law of energy conservation we have that
=> 
=> 
=>
