Answer:
The mass of the banana is m and it is at height h.
Applying the Law of Conservation of Energy
Total Energy before fall = Total Energy after fall
= 
Here, total energy is the sum of kinetic energy and potential energy
+ 
= 
+ 
(a)
When banana is at height h, it has
= 0 and = mgh
and when it reaches the river, it has
= 1/2m
and = 0
Putting the values in equation (a)
0 + mgh = 1/2m + 0
mgh = 1/2m
<em>cutting 'm' from both sides</em>
<em> </em>gh = 1/2
v = 
Hence, the velocity of banana before hitting the water is
v =
control is constant for given set of readings.
set indep var
observe dep var
0.36 J of work is done in stretching the spring from 15 cm to 18 cm.
To find the correct answer, we need to know about the work done to strech a string.
<h3>What is the work required to strech a string?</h3>
- Mathematically, the work done to strech a string is given as 1/2 ×K×x².
- K is the spring constant.
<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>
- The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
- Then K = Force / x.
- Here x = 15 - 10 = 5 cm = 0.05 m
- K = 40/0.05 = 800N/m.
<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>
- Work done = 1/2×k×x²
- Here x= 18-15=3cm or 0.03 m
- So, W= 1/2×800×0.03² = 0.36 J.
Thus, we can conclude that the work done is 0.36 J.
Learn more about the spring force here:
brainly.com/question/14970750
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A is the only logical answer
Answer:
- Newton's first law applies. An object at rest will stay that way until a force is applied.
- Any amount of effort can be applied to any amount of mass (in the ideal case). The question is not sufficiently specific.
Explanation:
A force is required to move an object because the object will stay at rest until a force is applied.
__
The effort required to lift or push two masses instead of one depends on the desired effect. For the same kinetic energy, no more effort is required. For the same momentum, half the effort is required for two masses. For the same velocity, double the effort is required.
