The answer for the following problem is explained below.
<u>Therefore the change in the momentum is 240 kg m/s</u>
Explanation:
Given:
net force() = 120 N
time (t) = 2 seconds
To solve:
change in momentum(Δp)
We know;
= Δp÷t
Δp = × t
Δp = 120 × 2
Δp =240 kg m/s
<u>Therefore the change in the momentum is 240 kg m/s</u>
Answer:
a) W_total = 8240 J
, b) W₁ / W₂ = 1.1
Explanation:
In this exercise you are asked to calculate the work that is defined by
W = F. dy
As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.
W = F dy = F Δy
let's apply this formula to our case
a) Let's use Newton's second law to calculate the force in the first y = 5 m
F - W = m a
W = mg
F = m (a + g)
F = 80 (1 + 9.8)
F = 864 N
The work of this force we will call it W1
We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)
F₂ - W = 0
F₂ = W
F₂ = 80 9.8
F₂ = 784 N
The work of this fura we will call them W2
The total work is
W_total = W₁ + W₂
W_total = (F + F₂) y
W_total = (864 + 784) 5
W_total = 8240 J
b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use
W₁ / W₂ = F y / F₂ y
W₁ / W₂ = 864/784
W₁ / W₂ = 1.1
Given Information:
Index of refraction of glass = n₂ = 1.42
Angle of reflection = θ₁
Angle of refraction in terms of angle of reflection = θ₂ = 0.5θ₁
Required Information:
Angle of refraction = θ₂ = ?
Answer:
Angle of refraction = θ₂ = 44.76°
Explanation:
We know from the Snell's law that
n₁sinθ₁ = n₂sinθ₂
Where n₁ is index of refraction of air and n₂ is index of refraction of glass, θ₁ is the angle of reflection and θ₂ is the angle of refraction.
Assuming n₁ = 1
1*sinθ₁ = 1.42*sinθ₂
since θ₂ = 0.5θ₁
1*sinθ₁ = 1.42*sin0.5θ₁
1/1.42 = sin0.5θ₁/sinθ₁
sin⁻¹(1/1.42) = sin⁻¹(sin0.5θ₁/sinθ₁)
44.76° = 0.5θ₁
θ₁ = 44.76°/0.5
θ₁ = 89.52°
So the angle of refraction is
θ₂ = 0.5(89.52°)
θ₂ = 44.76°
Answer:
Virgo will be the highest zodiac constellations at night.
Answer:
1.28 m, 14 m/s
Explanation:
At the maximum height, the velocity is 0.
Given:
a = -9.8 m/s²
v₀ = 5.00 m/s
v = 0 m/s
x₀ = 0 m
Find:
x
v² = v₀² + 2a(x - x₀)
(0 m/s)² = (5.00 m/s)² + 2(-9.8 m/s²) (x - 0 m)
x = 1.28 m
The maximum speed is at the bottom of the well.
Given:
a = -9.8 m/s²
v₀ = 5.00 m/s
x₀ = 0 m
x = -8.5 m
Find:
v
v² = v₀² + 2a(x - x₀)
v² = (5.00 m/s)² + 2(-9.8 m/s²) (-8.5 m - 0 m)
v = -13.8 m/s
Rounded to 2 sig-figs, the maximum speed is 14 m/s.