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
Answer:
The correct option is (B).
Explanation:
The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

It is mentioned that, an asteroid with an orbital period of 8 years. So,

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.
Larry Finkelstein, Norman Fischer, and Cassius Schwartz have been overlooked, in my opinion.
I think that the girl has greater tangential acceleration because she is closer to the center and the acceleration is greater there.
Answer:
v₀ₓ = 15 m / s,
= 5.2 m / s
v = 15.87 m / s
, θ = 19.1
Explanation:
This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed
Vy² =
² - 2 g y
² =
² + 2 g y
= √ (
² + 2 gy
Let's calculate
= √ (1.25² + 2 9.8 1.3)
= √ (27.04)
= 5.2 m / s
The initial speed can be calculated by the initial speed
v = √ v₀ₓ² +
²
v = RA (15² + 5.2²)
v = 15.87 m / s
We look for the angle with trigonometry
tan θ = voy / vox
θ = tan⁻¹ I'm going / vox
θ = tan⁻¹ 5.2 / 15
θ = 19.1
The answer is
v₀ₓ = 15 m / s
= 5.2 m / s