Please help quickly
his handbag exerts a force of 10 N on the hook. Which of the following values is a possible tension for each side of the cloth loop on the handbag? Each side of the loop exerts a vertical force only, not a horizontal force, on the rest of the handbag.
(Points : 1)
0 N
5 N
10 N
20 N
Question 5. 5. A picture is hanging from a wire whose two halves each make an angle of 45º with the horizontal x axis, as shown. The picture and frame weigh 12 N. Find the tension in each wire.
(Points : 1)
0 N
8.5 N
9.2 N
12 N
6 N
Question 6. 6. A picture is hanging from a wire whose two halves each make an angle of 45º with the horizontal x axis, as shown. The tension in each side of the wire is 17 N. Find the vertical force that the wire exerts on the picture hook supporting it.
(Points : 1)
17 N
0 N
24 N
34 N
Question 7. 7. Two farm workers are pushing a small boulder. One is pushing north with a force of 80.0 N, and the other is pushing east with a force of 90.0 N. The boulder does not move. What is the magnitude of the horizontal component of force that the ground exerts on the boulder?
(Points : 1)
0 N
80 N
90 N
120 N
10 N
5 N
8.5 N
0 N
For question 4, the possible tensions for each side of the cloth loop on the handbag can be determined by analyzing the forces acting on the hook. Since the handbag exerts a force of 10 N on the hook, the tension in each side of the loop must be equal to or greater than 10 N. Therefore, the possible values for the tension on each side of the cloth loop are 10 N, 20 N, or any value greater than 20 N.
For question 5, to find the tension in each wire, we need to use the concept of vector components. Since each wire makes an angle of 45º with the horizontal x-axis, the vertical component of tension will be equal to the weight of the picture and frame, which is 12 N. Therefore, the tension in each wire is 12 N.
For question 6, we need to calculate the vertical force that the wire exerts on the picture hook. Since the tension in each side of the wire is 17 N, and the wire makes an angle of 45º with the horizontal x-axis, the vertical force can be calculated using the formula:
Vertical Force = Tension * sin(angle)
Vertical Force = 17 N * sin(45º)
Vertical Force = 17 N * 0.7071
Vertical Force ≈ 12 N
Therefore, the vertical force that the wire exerts on the picture hook is approximately 12 N.
For question 7, we need to find the magnitude of the horizontal component of force that the ground exerts on the boulder. Since one farm worker is pushing north with a force of 80.0 N and the other is pushing east with a force of 90.0 N, the horizontal component of force can be calculated using the formula:
Horizontal Force = √(Force_North^2 + Force_East^2)
Horizontal Force = √(80.0 N^2 + 90.0 N^2)
Horizontal Force = √(6400 N^2 + 8100 N^2)
Horizontal Force = √(14500 N^2)
Horizontal Force ≈ 120 N
Therefore, the magnitude of the horizontal component of force that the ground exerts on the boulder is approximately 120 N.
For Question 4, the possible tension values for each side of the cloth loop on the handbag can be determined by considering the forces acting on the hook.
Since the handbag exerts a force of 10 N on the hook, the tension in the loop on one side must be equal to this force to keep the hook in equilibrium. Therefore, the tension in one side of the loop can be 10 N.
Since the handbag is not accelerating, the net force on the hook must be zero. This means that the tension in the other side of the loop must be equal in magnitude but opposite in direction to the tension in the first side. Therefore, the tension in the other side of the loop can also be 10 N.
So, the possible tension values for each side of the cloth loop on the handbag are 10 N.
For Question 5, to find the tension in each wire, we can use the concept of equilibrium. Since the picture and frame weigh 12 N, this force acts vertically downward. The tension in each wire will be equal to the weight of the picture and frame divided by the cosine of the angle each wire makes with the horizontal x-axis.
Since each wire makes an angle of 45º with the horizontal x-axis, the cosine of 45º is equal to 1/sqrt(2) or approximately 0.707.
Therefore, the tension in each wire will be 12 N / 0.707, which is approximately 16.97 N or rounded to the nearest tenth, 17 N.
So, the tension in each wire is approximately 17 N.
For Question 6, since the tension in each side of the wire is 17 N and the wire makes an angle of 45º with the horizontal x-axis, the vertical force that the wire exerts on the picture hook can be found by taking the vertical component of the tension.
The vertical component of the tension can be found by multiplying the tension by the sine of the angle, which is sin(45º) or 1/sqrt(2) or approximately 0.707.
Therefore, the vertical force that the wire exerts on the picture hook is approximately 17 N * 0.707, which is approximately 12.02 N or rounded to the nearest tenth, 12 N.
So, the vertical force that the wire exerts on the picture hook is approximately 12 N.
For Question 7, to find the magnitude of the horizontal component of force that the ground exerts on the boulder, we can use the concept of equilibrium. Since the boulder does not move, the net force in the horizontal direction must be zero.
The worker pushing north with a force of 80.0 N and the worker pushing east with a force of 90.0 N create a right-angled triangle. The horizontal component of force that the ground exerts on the boulder is equal to the sum of these horizontal forces.
Using the Pythagorean theorem, we can find the magnitude of the horizontal component of force as sqrt((80.0 N)^2 + (90.0 N)^2).
Calculating this gives us approximately 118.92 N, which can be rounded to the nearest tenth as 120 N.
So, the magnitude of the horizontal component of force that the ground exerts on the boulder is approximately 120 N.