Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :
where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy 
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Answer: True
A water pump
belong to a positive displacement pump that provides constant flow of water at
fixed speed, regardless of changes in the counter pressure. The two main types
of positive displacement pump are rotary pumps and reciprocating pumps.
Moreover, water
pump is a reciprocating positive displacement pump that have an expanding
cavity on the suction side and a decreasing cavity on the discharge side. In
water pumps, the liquid flows into the pumps as the cavity on the suction side
expands and then the liquid flows out of the discharge as the cavity collapses
providing water in a pail.
Answer:
a. The thickness of the wire is 2.5 mm.
b. The wire is 0.25 cm thick.
Explanation:
Number of turns of the wire = 10
The length of total turns = 25 mm
a. The thickness of the wire can be determined by;
thickness of the wire = 
= 
= 2.5 mm
Therefore, the wire is 2.5 mm thick.
b. To determine the thickness of the wire in centimetre;
10 mm = 1 cm
So that,
2.5 mm = x
x = 
= 0.25 cm
The wire is 0.25 cm thick.
