Answer:
h = 3.3 m (Look at the explanation below, please)
Explanation:
This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy
Kinetic energy = 
m
Plug in the numbers = (4.0)(
)
Solve = 2(64) = 128 J
Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.
Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)
Kinetic energy = Potential Energy
128 J = 39.2h
h = 3.26 m
h= 3.3 m (because of significant figures)
The total work done on the car is 784Joule.
<h3>What's the acceleration of the car?</h3>
- As per Newton's equation of motion, V= U+at
- U= initial velocity= 0 m/s
V= vinal velocity= 20m/s
t= time = 10s
a= acceleration
=> a= 20/10= 2m/s²
<h3>What's the distance covered by the car in 10 seconds?</h3>
- As per Newton's equation of motion,
V²-U² = 2aS
- S= distance covered by the car
- So, 20²-0=2×2×S=4S
=> 400= 4S
=> S= 400/4= 100m
<h3>What's the work done on the car due to frictional force?</h3>
Work done by frictional force= frictional force × distance
= (0.2×4×9.8)×100
= 784Joule
Thus, we can conclude that the work done on the car is 784Joule.
Learn more about the work done here:
brainly.com/question/25573309
#SPJ1
Answer:
-20.158ft-lb
Explanation:
Check the attached files for the explanation.
False. The nuclear energy is found within the nucleus. Electrons are located outside the nucleus.
Answer:
2 m/s
Explanation:
From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.
Let the 10.00 kg mass be 
and the 12.0 kg mass be 
. When they collide and stick, they have a combined mass of 
.
Momentum is given by 
. Set up the following equation:
, where 
is the desired final velocity of the masses.
Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.
Solving yields:
