Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
This is a poorly written question.
<span>Out of the choices listed, the first one is the only one that includes
a true statement ... the greater the mass of two objects, the
greater
the gravitational attraction is between them.</span>
-- Newton's law of universal gravitation doesn't "suggest" that. It states it ...
boldly and unequivocally.
-- The law doesn't refer to the "greatness" of the mass of the two objects.
It refers to the product of their masses.
-- It's true that the law of universal gravitation can be massaged and
manipulated to reveal the existence of gravitational planetary orbits.
But there's a lot more to it than simply the masses.
For example ... the gravitational force between two objects is inversely
proportional to
(the distance between the objects)² .
It turns out that IF that exponent were not precisely, exactly 2.000000... ,
then gravitational orbits could not exist.
Answer:
Wl = 1740 N
Explanation:
maximum lift weight unaided = force exerted (F) = 650 N
length of the wheelbarrow (L) = 1.4 m
weight of the wheelbarrow (w) = 80 N
distance of center of gravity of the wheel barrow from the wheel = 0.5 m
distance of center of gravity of the load from the wheel = 0.5 m
find the weight of the load (Wl)
from the diagram attached we can see that there is going to be a rotation about the axis A. let clockwise rotation be positive
ΣT = (F x 1.4) - ((Wl x 0.5) + (w x 0.5) = 0
(F x 1.4) = ((Wl x 0.5) + (w x 0.5)
Wl =
Wl =
Wl = 1740 N
Answer:
yes; yes
Explanation:
Phases of the moon refers to the shapes of the moon due to the lit part of it visible from the Earth. On a new moon day, the moon comes between the sun and the earth such that the lit portion is not visible from the Earth. On a full moon day, the earth comes between the sun and the moon and the whole lit part is visible.
When one would view the earth from the moon, the earth would also be visible as going through the phases. The order would be reversed. Understand this with the following example, On a new moon day, the Earth would be visible completely lit from the moon. So it will be full Earth day on the moon. On a full moon day, the lit side of the Earth would be completely away and hence, from the moon, new earth would be there.
The components that must be present for work to be considered is a force and a movement in the same direction as the force. In the basic definition of work, a magnitude and displacement that occurs in the same direction is what makes up work. Among the choices, the correct answer is the first one.
