A cheetah is walking at 5.0m/s when it sees a zebra 35m away. What acceleration would be required for the cheetah to reach 25.0
1 answer:
Answer:
Acceleration,
Explanation:
Given that,
Initial speed of the Cheetah, u = 5 m/s
Distance covered, s = 35 m
Final speed of the cheetah, v = 25 m/s
We need to find the acceleration required for the cheetah to reach its final speed. The formula is as follows :
So, the acceleration of the cheetah is . Hence, this is the required solution.
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Answer:
The correct option is;
Man doing most work
Box gaining the most energy
D Y Q
Explanation:
The given parameters of the question are;
The distance the box P is pushed by the Man X = 0
The force the Man X applies to the box P = x N
The distance the box Q is lifted by the Man Y = h > 0 meters
The minimum force the Man Y applies to the box Q = W, the weight of the box
Work done = Force × Distance
Energy gained = Potential energy + Kinetic Energy = (Mass × Gravity × Height = Weight × Height = W × h) + 1/2 × Mass × Velocity²
The final velocity of either box = 0 m/s (The boxes are at rest on the ground or the shelf)
Therefore, Kinetic energy = 0 J
The work done by Man X = 0 × x = 0 J
The energy gained by the box P = W × 0 = 0 J
The work done by Man Y = W × h = W·h J
The energy gained by the box P = W × h = W·h J
We have, the work done by the man Y = W·h J is more than the work done by the man X = 0 J
The energy gained by the box P = W·h J is more than the energy gained by the box Q = 0 J
Therefore, the correct option is D, Man doing the most work is Y, box gaining the most energy is Q.
When an object absorbs an amount of energy equal to Q, its temperature raises by 
following the formula
where m is the mass of the object and
is the specific heat capacity of the material.
In our problem, we have
, 
and 
, so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal:
where m is the mass of the object and
In our problem, we have