A) The kinetic energy of an object is given by:

where m is the mass of the object, and v its speed. For the lion in our problem, m=45 kg and v=14.2 m/s, so its kinetic energy is

b) the increase in gravitational potential energy of the lion is given by:

where g is the gravitational acceleration, and

is the increase in altitude of the lion. In this problem,

, so the increase in gravitational potential energy is

c) When the fox reaches the top of the tree, its gravitational potential energy is

As it jumps, its kinetic energy is

So the total mechanical energy of the fox as it jumps is
As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[ here a stands for acceleration]



Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[here S is the distance]
![= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m](https://tex.z-dn.net/?f=%3D%2042%2A5.5%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B-7.64%5D%20%5B5.5%5D%5E2%20m)
[ans]
Answer: 60mph
Explanation:
Given the following :
First leg travel:
Distance = 30 miles
Time of travel= 30 minutes = 0.5 hour
Second leg travel:
Distance = 60 miles
Time of travel = one hour
Average speed :
Speed = total Distance / time of travel
Total distance in miles = (30 + 60) miles = 90 miles
Total time of travel = 1 hour + 0.5 hour = 1.5 hours
Average speed = total distance traveled / total travel time
Average speed = 90 miles / 1.5 hours
Average speed = 60 miles / hour
= 60mph
A single magnetic field is shown.
Answer:
Hello your question has some missing parts attached below is the complete question
answer : 4 μm
Explanation:
since the scale bar works the same way as a scale on a map , each bar will therefore represent 1 μm and the mature parent cell's is about 4 times the labeled value hence the Mature parent cell diameter will approximately be : 4 μm