1.A hand exerciser utilizes a coiled spring. A force of 89.0 N is required to compress the spring by 0.0191 m. Determine the force needed to compress the spring by 0.0508 m.

236.71 N

Explanation:

Applying,

F = ke Equation 1

Where F = force, k = spring constant, e = extension.

make k the subject of the equation

k = F/e Equation 2

From the question,

Given: F = 89 N, e = 0.0191 m

Susbtitute these values into equation 2

k = 89/0.0191

k = 4659.69 N/m

When e = 0.0508 m

The force needed is given as

F = (4659.69×0.0508)

F = 236.71 N

Force needed to compress spring by 0.0508 m = 236.71 N (Approx.)

Explanation:

Given:

Force applied = 89 N

Length of spring compress = 0.0191 m

Find:

Force needed to compress spring by 0.0508 m

Computation:

Using Hooke's law

F = K(x)

Constant K = F / x

Constant K = 89 / 0.0191

Constant K = 4,659.68 N/m

Force needed to compress spring by 0.0508 m

So,

Force needed to compress spring by 0.0508 m = K(X)

Force needed to compress spring by 0.0508 m = (4,659.68)(0.0508)

Force needed to compress spring by 0.0508 m = 236.71 N (Approx.)

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: