Answer:
I believe the answer to be B
Explanation:
If you were playing on grass, the ball would be able to roll around much easier rather than it to be on sand. If it's wrong I am so sorry
Answer:
2156 J
Explanation:
From the question,
Work done = Combined mass of the bucket and water×height×gravity.
W = (M+m)hg............................. Equation 1
Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.
Given: M = 20 kg, m = 2 kg, h = 10 m
Constant: g = 9.8 m/s²
Substitute these value into equation 1
W = (20+2)×10×9.8
W = 22×98
W = 2156 J
No work is done because the object needs to be moved. The formula for work is Work = Force x Distance.
Answer:
F₃ = 122.88 N
θ₃ = 20.63°
Explanation:
First we find the components of F₁:
For x-component:
F₁ₓ = F₁ Cos θ₁
F₁ₓ = (50 N) Cos 60°
F₁ₓ = 25 N
For y-component:
F₁y = F₁ Sin θ₁
F₁y = (50 N) Sin 60°
F₁y = 43.3 N
Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:
F₂ₓ = F₂ = 90 N
F₂y = 0 N
Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:
F₁ₓ + F₂ₓ + F₃ₓ = 0 N
25 N + 90 N + F₃ₓ = 0 N
F₃ₓ = - 115 N
for y-components:
F₁y + F₂y + F₃y = 0 N
43.3 N + 0 N + F₃y = 0 N
F₃y = - 43.3 N
Now, the magnitude of F₃ can be found as:
F₃ = √F₃ₓ² + F₃y²
F₃ = √[(- 115 N)² + (- 43.3 N)²]
<u>F₃ = 122.88 N</u>
and the direction is given as:
θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)
<u>θ₃ = 20.63°</u>