A rectangular loop of wire that can rotate about an axis through its center is placed between the poles of a magnet in a magneti
c field with a strength of 0.40 T. The length of the loop L is 0.16 m and its width w is 0.040 m. What is the magnetic flux through the loop when the plane of the loop is perpendicular to the magnetic field?
1 answer:
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Refer to the diagram shown below.
The given data is
mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)
Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.
Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m
8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m
Answer: (1.5, 0.5) m
The given data is
mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)
Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.
Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m
8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m
Answer: (1.5, 0.5) m
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
- ∑Fᵧ = maᵧ
- T₁ - m₂g = m₂aᵧ
Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
- T₁ = m₂aᵧ + m₂g
- T₁ = m₂(a + g)
We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
- ∑Fₓ = maₓ
- F_f - F_g sinΘ = maₓ
The normal force is equal to the x-component of the force of gravity.
- (F_n · μ_k) - m₁g sinΘ = m₁aₓ
- (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
- [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
- [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
- [2.539595871] - [-58.0962595] = 6aₓ
- 60.63585537 = 6aₓ
- aₓ = 10.1059759 m/s²
Now let's go back to this equation:
- T₁ = m₂(a + g)
We have 3 known variables and we can solve for the tension force.
- T = 2(10.1059759 + 9.8)
- T = 2(19.9059759)
- T = 39.8119518 N
The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.
