Answer:
The maximum torque in the coil is 
.
Explanation:
Given that,
Number of turns in the circular coil, N = 50
Radius of coil, r = 5 cm
Magnetic field, B = 0.5 T
Current in coil, I = 25 mA
We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :
For maximum torque, 
So, the maximum torque in the coil is .
Answer:
Answer:
New speed of the 22-kg block is 1.57 m/s
Explanation:
Mass of block
Mass of another block
Initial speed of the block
Initial speed of the another block
Initial speed of the another block
For conservation of momentum, we have
Substitute all the values and solving for final speed of the 22kg block is
new speed of the 22-kg block is 1.57 m/s
Couldnt write the answer so check picture
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0.2 is the value of coefficient of friction (k)
F=kN
F=horizontal force
n=Normal Force
k=coefficient of friction
k=F/N
k=200/1000
k=0.2
The ratio of the normal force pushing two surfaces together to the frictional force preventing motion between them is known as the friction coefficient. Usually, the Greek letter mu is used to indicate it .N is the normal force, and F is the frictional force, hence F = N/N.
Due to the fact that both F and N are measured in units of force, the coefficient of friction has no dimensions (such as newtons or pounds). The coefficient of friction can have a variety of values for both static and dynamic friction. Static friction occurs when an object encounters friction that resists any applied force, keeping the object at rest until the static frictional force is released. In kinetic friction, the frictional force resists the motion of the object.
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Answer:
Explanation:
Given
Required
Determine the impulse
The impulse is calculated as follows:
Substitute values for Force and Time
<em>Hence, the impulse experienced is 8.0Ns</em>
