Answer:
The induced emf in the coil is 0.522 volts.
Explanation:
Given that,
Radius of the circular loop, r = 9.65 cm
It is placed with its plane perpendicular to a uniform 1.14 T magnetic field.
The radius of the loop starts to shrink at an instantaneous rate of 75.6 cm/s , 
Due to the shrinking of radius of the loop, an emf induced in it. It is given by :
So, the induced emf in the coil is 0.522 volts.
Answer:
The canon B hits the ground fast.
Explanation:
Given that,
Speed of cannon A = 85 m/s
Speed of cannon B= 100 m/s
Speed of cannon C = 75 m/s
We need to calculate the cannonballs will hit the ground with the greatest speed
Using conservation of energy
The final kinetic energy of canon depends on initial kinetic energy and potential energy.
The final velocity depends upon initial velocity and initial height.
So, the initial velocity of canon B is high.
Hence, The canon B hits the ground fast.
Answer:
1.876 J
Explanation:
First, let’s calculate the compression of the spring from the Hooke’s law:
F=kx,
here, F=75 N is the force acted on the spring, k=1500 N⁄m is the force constant of the spring, x is the compression of the spring.
Then, we get:
x=F/k=(75 N)/(1500 N/m)=0.05 m.
Finally, we can find the potential energy stored in the spring:
PE=1/2 kx^2=1/2∙1500 N/m∙(0.05 m)^2=1.875 J.
correct my answer if it's wrong ^^
Answer:
C
Explanation:
b is 55 miles per hour south or 55mph[South]
the important part to understand is that velocity is a vector meaning it also has to have a direction, in this case south. That is why a (2m/s) is not a velocity, however 2m/s is a speed.
Answer:
0.3cm
Explanation:
Y = 0.3 sin(0.5x - 50t) compare with,
y = A sin(kx - wt)
A = 0.3m
