Answer:
10miles/second
Explanation:
Change in velocity of the bird is expressed as the difference between the final velocity and initial velocity of the body.
Change in velocity = Final velocity - initial velocity
Since the bird takes off from the tree, the initial velocity of the bird = 0miles/sec
Final velocity = 10miles/secs
Change in velocity = 10-0
Change in velocity = 10miles/second
If you're moving, then you have kinetic energy.
If you're not at the bottom yet, then you still have
some potential energy left.
Answer:
a.
Explanation:
cuz whenever the ball was traveling up the staircase it was building up energy to use (potential energy) unlike b.
<span>Kinetic energy because it is taking the students to school.</span>
Answer:
The value is 
Explanation:
From the question we are told that
The initial speed is 
Generally the total energy possessed by the space probe when on earth is mathematically represented as

Here
is the kinetic energy of the space probe due to its initial speed which is mathematically represented as
=>
=> 
And
is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

Here
is the escape velocity from earth which has a value 
=> 
=> 
Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

Generally from the law energy conservation we have that
So

=> 
=> 
=> 