Answer:
<u>FALSE.</u>
Explanation:
Newton's third law states that :
- <em>Every action has equal and opposite reaction</em>
- <em>That is , the magnitude is the same but the directions are opposite</em>
- <em>The action reaction forces DONOT operate on the same body.</em>
For example ,
If a block is kept on the ground , the action force is the normal force acting on it due to the ground. <em>BUT , NOTE THAT : the reaction force isn't the gravitational force on the body ! It is the normal force acting on the ground due to the block !</em>
Thus,
we conclude that action and reaction forces donot act on the same body and therefore , this case has the <u>answer : FALSE </u>
Answer:
a= - 1.2 m/s²
Direction of a : in the negative x- direction
Explanation:
Because the ball moves with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (m)
v₀: initial velocity in m/s
vf: final velocity in m/s
a: acceleration in m/s²
Known data
v₀ = 9.0 m/s
vf= 3.00 m/s
d = 30.0 m.
Problem development
We replace data in the formula (1) to calculate acceleration :
vf²=v₀²+2*a*d
(3)²= (9)² + (2* 30)*a
9=81+(60)*a
9-81=(60)*a
-72= (60)*a
a= -72/60
a= - 1.2 m/s²
Direction of a : in the negative x- direction
Answer:
a)speed of light.
a)speed of light×time=distance.
Explanation:
light has a constant speed of 299,792,458m/s
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