Kinetic energy = (1/2) (mass) x (speed)²
At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules
At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules
The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.
That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
Electrons are transferred sequentially between the two photosystems, with photosystem I acting to generate NADPH and photosystem II acting to generate ATP. The pathway of electron flow starts at photosystem II, which is homologous to the photosynthetic reaction center of R. viridis already described.
Answer:
The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.
Explanation:
Given that,
Mass of aircraft = 10000 kg
Speed = 620 km/h = 172.22 m/s
Altitude = 10 km = 1000 m
We calculate the change in potential energy
For g = 10 m/s²,
The change in potential energy will be 1000 MJ.
We calculate the change in kinetic energy
For g = 10 m/s²,
The change in kinetic energy will be 150 MJ.
Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.
The image of the object is 8cm to the left of the lens (D)
<h3>
</h3>
What is the image of an object?
The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.
It is calculated thus:
1÷v = 1÷f - 1÷u
<h3>How to calculate the image of an object</h3>
From the formula
1÷v = 1÷f - 1÷u
<h3>
Where </h3>
V = image distance fromthe object
U = object
f = focal length
Substitute the values
1÷v = 1÷8 - 1÷ 4
1÷v = - 1÷8
Make v the subject of formula
v = -8cm
Therefore, the image of the object is 8cm to the left of the lens (D)
Learn more on focal length here:
brainly.com/question/25779311
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