Apiece of steel is held firmly in the jaws of a vise. a force larger than 3350 n will cause the piece of steel to start to move out of the vise. if the coefficient of friction between the steel and each of the jaws of the vise is 0.825 and each jaw applies an equal force, what is the magnitude of the normal force exerted on the steel by each jaw?

N=2030.30 N

Explanation:

Given that

F= 3350 N

Coefficient of friction ,μ =0.825

Lets take N is the normal force

We know that friction force fr

fr= μ N

The total force in left side = 2 fr

Total force in right side =F

So by equating the above two forces

F= 2 fr

3350 = 2 x 0.825 x N

N=2030.30 N

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: