Coal is lifted out of a mine a vertical distance of 50 m by an engine that supplies 500 w to a conveyer belt. how much coal per minute can be brought to the surface? ignore the effects of friction.

61 kg

Explanation:

Given data

distance (d): 50 m

power (P): 500 W = 500 J/s

mass (m): ?

time (t): 1 min = 60 s

The engine must do some work (w) to lift the coal. The work done in 1 minute is:

P = w / t

w = P . t = 500 J/s × 60 s = 3.0 × 10⁴ J

The work is equal to the force exerted (F) times the distance (d).

w = F × d

F = w / d = 3.0 × 10⁴ J / 50 m = 6.0 × 10² N

The force exerted is equal to the mass lifted (m) times the acceleration. Here, the acceleration to surpass is that of gravity (g = 9.8 m/s²).

F = m × g

m = F / g = 6.0 × 10² N / 9.8 m/s² = 61 kg

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:

3. dark matter

empty space could also be a simple vacuum