Acceleration is a measure of how fast the velocity of a particle/object is changing at a point. Therefore, acceleration can be measured by finding the slope of the velocity of the particle at a certain point.
Since we are given the graph of velocity, we have to find the slope of the particle at point B. In the given graph, the slope of the velocity of the particle at point B is zero. Therefore, the acceleration of the particle at point B is 0.
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Answer:
v_f = 1.05 m/s
Explanation:
From conservation of energy;
E_f = E_i
Thus,
(1/2)m(v_f)² + (1/2)I(ω_f)² + m•g•h_f + (1/2)k•(x_f)² = (1/2)m(v_i)² + (1/2)I(ω_i)² + m•g•h_i + (1/2)k•(x_i)²
This reduces to;
(1/2)m(v_f)² + (1/2)Ik(x_f)² = (1/2)k•(x_i)²
Making v_f the subject, we have;
v_f = [√(k/m)] * [√((x_i)² - (x_f)²)]
We know that ω = √(k/m)
Thus,
v_f = ω[√((x_i)² - (x_f)²)]
Plugging in the relevant values to obtain;
v_f = 17.8[√((0.068)² - (0.034)²)]
v_f = 17.8[0.059] = 1.05 m/s
Answer:
change in gravitational potential energy Δ PE = 392 J
Explanation:
given data
mass of the person m = 80 kg
height of the centre of mass Δh = 0.50 m
to find out
change in gravitational potential energy
solution
we get here change in gravitational potential energy that is express here as
change in gravitational potential energy Δ PE = m × g × Δh .........1
put here value we get
change in gravitational potential energy Δ PE = m × g × Δh
change in gravitational potential energy Δ PE = 80 × 9.8 × 0.50
change in gravitational potential energy Δ PE = 392 J
Answer:
3.33 N
Explanation:
First, find the acceleration.
Given:
Δx = 3 m
v₀ = 0 m/s
t = 3 s
Find: a
Δx = v₀ t + ½ at²
3 m = (0 m/s) (3 s) + ½ a (3 s)²
a = ⅔ m/s²
Use Newton's second law to find the force.
F = ma
F = (5 kg) (⅔ m/s²)
F ≈ 3.33 N
Answer:
1. F = M x A
2. Force
3. 2nd Law: Force
4. a, b, c (in order)
5. 3rd Law: Action and Reaction
6. b, c, a (in order)
7. 1st Law: Inertia