8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.
1 answer:
Answer:
The resultant tension of the two ropes is approximately 42.4 N
The length of the line representing the resultant tension is approximately 8.48 cm
Please find included with the answer the scale drawing created with Microsoft Word
Explanation:
The given parameters are;
The tension in rope P, = 30 N
The tension in rope Q, = 30 N
The angle the rope, 'P', makes with the horizontal = 45°
The angle the rope, 'Q', makes with the horizontal = 45°
The scale factor of the scale diagram, S.F. = 5.0 N/cm
By the resolution of forces at equilibrium, we have;
The sum of the vertical forces, =
+
+ W = 0
∴ W = -( +
)
W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712
The weight of the heavy ball, W ≈ 42.4 N acting downwards
The sum of the horizontal forces, =
+
= 0
The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm
The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included
You might be interested in
Answer:
The hottest temperature is
Explanation:
From the question we are given
Generally converting to Fahrenheit
=>
=>
Converting to Fahrenheit
=>
=>
Now comparing the temperature in Fahrenheit we see that is the hottest
Answer:
<em>113.4 J</em>
Explanation:
<u>Elastic Potential Energy</u>
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.
The spring has a natural length of 0.7 m and a spring constant of k=70 N/m. When the spring is stretched to a length of 2.5 m, the change of length is
Δx = 2.5 m - 0.7 m = 1.8 m
The energy stored in the spring is:
PE = 113.4 J
Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;
Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;
The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N