The tension in the cable is 23.2 N
<h3>What is the tension in the string?</h3>
The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.
Tcosθ, is acting perpendicularly, Tcosθ = 0
Taking moments about the pivot:
Tsinθ * 2.2 = 4 * 9.8 * 0.7
Solving for θ;
θ = tan⁻¹(1.4/2.2) = 32.5°
T = 27.44/(sin 32.5 * 2.2)
T = 23.2 N
In conclusion, the tension in the cable is determined by taking moments about the pivot.
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Explanation:
V=40m/s
Vy=V.sina=40.sin20=40 . 0.342=13.68m/s
Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s
Projectile travels during 5 seconds and the ramge becomes:
x=V.t=30.64 . 5=153.2m
Answer:
The force per unit length (N/m) on the top wire is 16.842 N/m
Explanation:
Given;
distance between the two parallel wire, d = 38 cm = 0.38 m
current in the first wire, I₁ = 4.0 kA
current in the second wire, I₂ = 8.0 kA
Force per unit length, between two parallel wires is given as;

where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
Substitute the given values in the above equation and calculate the force per unit length

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m
Answer:
The magnitude of the acceleration is equal to 19.6m/s² and the acceleration is directed upwards though the magnitude of the charge has doubled. This is because the electric force is directed upwards and from newton's second law of motion the charge will have acceleration in the same direction as the electric force on the charge.
Explanation:
The detailed solution can be found in the attachment below.
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Answer:
12m
Explanation:
Given parameters:
Distance walked westward = 2.8m
Time of travel = 5min
Distance walked eastward = 9.2m
Time of travel = 10min
Unknown:
The total shopper's travel distance = ?
Solution:
Total distance traveled is the sum of the length of path covered by a body. It is a scalar quantity.
Total distance = distance walked westward + distance walked eastward
Total distance = 2.8m + 9.2m = 12m