Answer:
4000J
Explanation:
Given parameters:
Weight of the man = 800N
Height of ladder = 5m
Unknown:
Gravitational potential energy gained = ?
Solution:
The gravitational potential energy is due to the position of a body.
Gravitational potential energy = weight x height
Now insert the parameters;
Gravitational potential energy = 800 x 5 = 4000J
The human population which grows by about 2 percent each year, is showing linear growth.
<h3>What is linear increment?</h3>
A linear increment occurs when a sequence increases gradually with a constant value.
If a population increases at the rate of 2%, the incremental rate or constant factor is 2%, thus, we can conclude that the human population which grows by about 2 percent each year, is showing linear growth.
Learn more about linear increment here: brainly.com/question/2511791
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Answer:
t = 3.38 s
Explanation:
We have,
Initial speed of the ball that leaves the student's hand is 16 m/s
Initially, the hand is 1.90 m above the ground.
It is required to find the time for which the ball in the air before it hits the ground. We can use the equation of kinematics as :
Here, and a=-g
The equation become:
After rearranging we get the above equation as :
It is a quadratic equation, we need to find the value of t. On solving the above equation, we get :
t = -0.115 s and t = 3.38 s (ignore t = -0.115 s )
So, the ball is in air for 3.38 seconds before it hits the ground.
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