A box is sliding on a rough patch of ice, and the only horizontal force acting on it is friction. You observe that it slides to a stop in 600 seconds over a distance of 23 meters. What magnitude coefficient of kinetic friction between the box and the ice?

The magnitude of the coefficient of kinetic friction is approximately 1.304 × 10⁻⁵

Explanation:

The given parameters are;

The time it takes for the box to stop, "t" = 600 seconds

The distance the box moves before coming to a stop, "d" = 23 meters

The work done by the friction force to stop the block = The kinetic energy lost by the box

The work done by the friction force to stop the block = ·m·g·d

The kinetic energy change of the block = 1/2·m·u²

Where;

g = The acceleration due to gravity = 9.8 m/s²

m = The mass of the box

= The coefficient of kinetic friction

d = The distance travelled by the block = 600 m

u = The initial velocity of the block

∴ ·m·g·d = 1/2·m·u²

·g·d = 1/2·u²

= 1/2·u²/(g·d) = 1/2·u²/(9.8 × 23) = u²/450.8

∴   = u²/450.8...(1)

The kinematic equation of motion of the block can be presented as follows;

v = u - a·t

Where;

v = The final velocity = 0 m/s (The block slides to a stop)

u = The initial velocity

a = The acceleration of the block (the minus sign indicates that the block's velocity is decreasing)

t = The time of motion of the block = 600 seconds

Therefore, by substituting the above values, we have;

0 = u - 600·a

∴ u = a·600

The kinematic equation of motion for the distance moved by the block, "s" can be expressed as follows;

s = u·t - 1/2·a·t²

Where;

s = The distance moved by the bloc = 23 meters

∴ 23 = a·600·600 - 1/2·a·600² = 360,000·a - 1/2·a·360,000 = 180,000·a

a = 23/(180,000)

u = 600 × (23/(180,000)) = 23/300

The initial velocity, u = 23/300 m/s

From equation (1), we have;

= u²/(450.8) = (23/300)²/450.8 = 23/1764000 ≈ 1.304 × 10⁻⁵

The magnitude of the coefficient of kinetic friction, ≈ 1.304 × 10⁻⁵

d they tend to lose electrons and become cations.

explanation:

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: