Brainliest if right.
1 answer:
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Answer and Explanation:
(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope (), vertical gravitational force (
) and vertical normal force (
), due to the surface. Since there is no vertical movement,
and
cancels it out. So, for this block, net force is horizontal due to the rope
.
The block of mass m is hanging from the pulley, so there is the force of the rope () and the gravitational force (
). Both are vertical, because there is no surface "holding" block m.
(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:
a = 5.22
The acceleration of either block is 5.22 m/s².
(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².
Suppose positive referential is going up. To determine mass:
m = 1.25
Block m has 1.25 kg.
(d) Gravitational force is also called weight. So, as described above: .
The weight for the hanging block is
12.25 N
Comparing tension and weight:
≈ 0.65
We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.
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