Answer:
W = 2352 J
Explanation:
Given that:
- mass of the bucket, M = 10 kg
- velocity of pulling the bucket, v = 3
- height of the platform, h = 30 m
- rate of loss of water-mass, m =
Here, according to the given situation the bucket moves at the rate,
The mass varies with the time as,
Consider the time interval between t and t + ∆t. During this time the bucket moves a distance
∆x = 3∆t meters
So, during this interval change in work done,
∆W = m.g∆x
<u>For work calculation:</u>
![W=\int_{0}^{10} [(10-0.4t).g\times 3] dt](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7B10%7D%20%5B%2810-0.4t%29.g%5Ctimes%203%5D%20dt)
![W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}](https://tex.z-dn.net/?f=W%3D%203%5Ctimes%209.8%5Ctimes%20%5B10t-%5Cfrac%7B0.4t%5E%7B2%7D%7D%7B2%7D%5D%5E%7B10%7D_%7B0%7D)
1. D - sound travels the fastest through solids
2. 50 mm/s - v=fa
3. B - only process that involves changing waves
Answer:
wrong question
Explanation:
unit of acceleration is written in m/s which is wrong
Electricity can travel in a closed circuit.
The maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee 119.28N.
Given,
m=76kg, μ=0.16, g=9.81 m/
frictional force = mμg= 76*0.16*9.81= 119.28 N
<h3>Frictional force</h3>
An opposing force to the relative motion of two bodies in contact is known as frictional force. Always acting in the opposite direction from the direction of motion, frictional force is applied to a moving body. Because it resists motion, it aids in lowering the moving object's speed. The force is one of touch. Four broad categories can be made for the force of friction depending on the sort of motion that occurs between the two objects. Static friction is the force of friction between an object and the surface it is put on.
What is the maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee?
Learn more about frictional force here:
brainly.com/question/13707283
#SPJ4
