Answer:
v = 5.34[m/s]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.
Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.
E₁ = mechanical energy at initial state [J]
In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.
In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.
E₂ = mechanical energy at final state [J]
Now we can use the first statement to get the first equation:
where:
W₁₋₂ = work from the state 1 to 2.
where:
h = elevation = 1.5 [m]
g = gravity acceleration = 9.81 [m/s²]
![58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]](https://tex.z-dn.net/?f=58%20%3D%20v%5E%7B2%7D%20%2B29.43%5C%5Cv%5E%7B2%7D%20%3D28.57%5C%5Cv%3D%5Csqrt%7B28.57%7D%5C%5Cv%3D5.34%5Bm%2Fs%5D)
Differentiation in its simplest of terms means breaking something into small parts. On the other hand, integration is taking those really small parts and gluing them in the right order. In short, these terms are the direct opposite or inverses of each other. The term which can tell you how fast you are going at a moment in time at ones current location is called a derivative. The term on the other hand, which can tell you how far you have travelled if you have been keeping track of your location and your time is what an integral is referred to. It is like differentiation only needs knowledge on the local neighbourhood while integration will need the knowledge on a global knowledge.
vib. motion motion of wire of guitar
circular motion revolution of earth around sun
1ml 1cm3
1m3 100cm3
volume of liquid measuring cylinder
Answer:
Explanation:
Given
mass of box 
speed of box 
distance moved by the box 
coefficient of kinetic friction 
Friction force 
Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy
This is a great problem if you like getting tied up in knots
and making smoke come out of your brain.
I found that it makes the problem a lot easier if I give the objects some
numbers. I'm going to say that the mass of Object 5 is 20 clods.
Let the mass of Mass of Object 5 be 20 clods .
Then . . .
-- The mass of Object 2 is double the mass of Object 5 = 40 clods.
-- The mass of Object 4 is half of the mass of Object 5 = 10 clods.
and
-- the mass of Object 3 is half of the mass of Object 4 = 5 clods.
So now, here are the masses:
Object #1 . . . . . unknown
Object #2 . . . . . 40 clods
Object #3 . . . . . 5 clods
Object #4 . . . . . 10 clods
Object #5 . . . . . 20 clods .
Now let's check out the statements, and see how they stack up:
Choice-A:
Object 3 and Object 5 exert the same gravitational force on Object 1.
Can't be.
Objects #3 and #5 have different masses, so they can't both
exert the same force on the same mass.
Choice-B.
Object 2 and Object 4 exert the same gravitational force on Object 1.
Can't be.
Objects #2 and #4 have different masses, so they can't both
exert the same force on the same mass.
Choice-C.
The gravitational force between Object 1 and Object 2 is greater than
the gravitational force between Object 1 and Object 4.
Yes ! Yay !
Object-2 has more mass than Object-4 has, so it must exert more force on
ANYTHING than Object-4 does, (as long as the distances are the same).
Choice-D.
The gravitational force between Object 1 and Object 3 is greater than the gravitational force between Object 1 and Object 5.
Can't be.
Object-3 has less mass than Object-5 has, so it must exert less force on
ANYTHING than Object-4 does, (as long as the distances are the same).
Conclusion:
If the DISTANCE is the same for all the tests, then Choice-C is
the only one that can be true.
