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<u>Answer: Koala bears are considered herbivores, or as in the scientific name, arboreal herbivorous marsupial, marsupial because it also carries it's babies around in a pouch. Koala bears are also native to Australia, which eucalyptus leaves are also native to.
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b). The power depends on the RATE at which work is done.
Power = (Work or Energy) / (time)
So to calculate it, you have to know how much work is done AND how much time that takes.
In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A. But the question doesn't tell us anywhere how much time that takes. So there's NO WAY to calculate the power needed to do it.
The more power is used, the faster the car is lifted. The less power is used, the slower the car creeps up the first hill. If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill. That would just take a long time, but it could be done if the power is small enough.
Without knowing the time, we can't calculate the power.
...
d). Kinetic energy = (1/2) · (mass) · (speed squared)
On the way up, the car stops when it reaches point-A.
On the way down, the car leaves point-A from "rest".
WHILE it's at point-A, it has <u><em>no speed</em></u>. So it has no (<em>zero</em>) kinetic energy.
Answer:
Option (c).
Explanation:
Given that,
Mass of a cart, m = 0.8 kg
Tension in the car, F = 4 N
Net force, F = ma
a is acceleration of the cart.
So,
So, the acceleration of the cart is 
.
Answer:
20.96 m/s
Explanation:
Using the equations of motion
y = uᵧt + gt²/2
Since the puck slides off horizontally,
uᵧ = vertical component of the initial velocity of the puck = 0 m/s
y = vertical height of the platform = 2 m
g = 9.8 m/s²
t = time of flight of the puck = ?
2 = (0)(t) + 9.8 t²/2
4.9t² = 2
t = 0.639 s
For the horizontal component of the motion
x = uₓt + gt²/2
x = horizontal distance covered by the puck
uₓ = horizontal component of the initial velocity = 20 m/s
g = 0 m/s² as there's no acceleration component in the x-direction
t = 0.639 s
x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m
For the final velocity, we'll calculate the horizontal and vertical components
vₓ² = uₓ² + 2gx
g = 0 m/s²
vₓ = uₓ = 20 m/s
Vertical component
vᵧ² = uᵧ² + 2gy
vᵧ² = 0 + 2×9.8×2
vᵧ = 6.26 m/s
vₓ = 20 m/s, vᵧ = 6.26 m/s
Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s
You can do this two ways:
1). Whatever kinetic energy the rolling ball has is the amount
of energy you have to absorb in order to stop it.
2). Whatever momentum the rolling ball has is the amount of
momentum you have to provide in the other direction to cancel it.
Since you asked about force and time, we sense 'impulse' in the
air, and we know that impulse is exactly a change in momentum.
So let's use #2 and talk about momentum and impulse.
Impulse = (force) x (time)
Momentum of a moving object is (mass) x (speed) .
-- Momentum of the first ball: (8 kg) x (0.2 m/s) = 1.6 kg-m/s
Impulse required to stop it = 1.6 kg-m/s
(force) x (10 sec) = 1.6
Force required = 1.6 / 10 = 0.16 Newton .
-- Momentum of the second ball: (4 kg) x (1 m/s) = 4 kg-m/s
Impulse required to stop it = 4 kg-m/s
(force) x (10 sec) = 4
Force required = 4 / 10 = 0.4 Newton .
You need more force o stop the second ball. Although its mass
is only 1/2 the mass of the 8kg ball, it's moving 5 times as fast,
and has 2.5 times the momentum of the bigger ball.
So you need 2.5 times as much impulse to stop it.
If you're going to push on each ball for the same length of time,
then you need to push 2.5 times as hard on the smaller ball in
order to stop it.
