Answer:
1.53 s
Explanation:
Initially vertical component of velocity of the ball, uy = 7.5 m/s
Net displacement is vertical direction is zero, Δy =0
Use second equation of motion:
Δy = uy t + 0.5 a t²
Here, acceleration a = -g (g =9.8 m/s²)
Substitute all the values and solve for g
0 = 7.5 t -0.5 (9.8)t²
7.5 t = 4.9 t²
t = 1.53 s
Since it is a smooth marble and basing from the graph, it has a constant slope, you could just get any two points along the line and get its slope since speed is Δd/Δt. Let the points be (4,6) and (6,9). Thus,
Δd/Δt = (9-6)/(6-4) = 1.5
Thus, the average speed is 1.5 m/s.
Answer:
Trial 1 is the largest, trial 3 is the smallest
Explanation:
Given:
<em>Trial 1</em>
M₁ = 6·10²² kg
d₁ = 3 500 km = 3.5·10⁶ м
<em>Trial 2</em>
M₂ = 6·10²² kg
d₂ = 7 000 km = 7·10⁶ м
<em>Trial 3</em>
M₃ = 3·10²² kg
d₃ = 7 000 km = 7·10⁶ м
___________
F - ?
Gravitational force:
F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)
F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)
F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)
Trial 1 is the largest, trial 3 is the smallest
Answer:
27000 Nm
Explanation:
The boom end at A is fixed and end at B is subjected to a 3kN force. Boom AB has length of 9m. The moment about A is the product of force 3kN (3000 N) at B and the moment arm of 9m
M = FL = 3000 * 9 = 27000 Nm
So the moment about A is 27000 Nm