The work done on the filled bucket in raising out of the hole is 2, 925 Joules
<h3>How to determine the work done</h3>
Using the formula:
Work done = force * distance
Note that force = mass * acceleration
F = mg + ma
F = 4. 5 * 10 + 28 * 10
F = 45 + 280
F = 325 Newton
Distance = 9m
Substitute into formula
Work done = 325 * 9
Work done = 2, 925 Joules
Therefore, the work done is 2, 925 Joules
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Answer:
![W=2.76\times 10^{-6}\ J](https://tex.z-dn.net/?f=W%3D2.76%5Ctimes%2010%5E%7B-6%7D%5C%20J)
Explanation:
Given that,
The distance between two spheres, r = 25 cm = 0.25 m
The capacitance, C = 26 pF = 26×10⁻¹² F
Charge, Q = 12 nC = 12 × 10⁻⁹ C
We need to find the work done in moving the charge. We know that, work done is given by :
![U=\dfrac{Q^2}{2C}](https://tex.z-dn.net/?f=U%3D%5Cdfrac%7BQ%5E2%7D%7B2C%7D)
Put all the values,
![U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J](https://tex.z-dn.net/?f=U%3D%5Cdfrac%7B%2812%5Ctimes%2010%5E%7B-9%7D%29%5E2%7D%7B2%5Ctimes%2026%5Ctimes%2010%5E%7B-12%7D%7D%5C%5C%5C%5CU%3D2.76%5Ctimes%2010%5E%7B-6%7D%5C%20J)
So, the work done is
.
A. 1.35 is the number in between 1.2 and 1.5.
Answer:
Speed: Distance per time, 400 km/h, and a scalar quantity.
Velocity: Displacement per time, 20 m/s south, and a vector quantity.
Explanation:
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A large force is required to accelerate the mass of the bicycle and rider. Once the desired constant velocity is reached, a much smaller force is sufficient to overcome the ever-present frictional forces.