Equation for ke = 1/2mv^2
1) ke = 1/2 x 5000 x (2x2)
= 10,000J
2) ke = 1/2 x 4000 x (3x3)
= 18,000J
So Object 2 has more Kinetic Energy
A :-) A.a.) Given - u = 0.00 m/s
v = 10.0 m/s
t = 1.5 sec
m = 75 kg
Solution -
a = v - u by t
a = 10 - 0 by 1.5
a = 10 by 1.5
a = 6.6 m/s^2
A.b.) sorry ! I don’t no how to do this question
Answer:
1800J
Explanation:
Given parameters:
Weight of the book = 20N
Total distance covered = 45m + 15m + 30m = 90m
Unknown:
Total work performed on the books = ?
Solution:
To solve this problem we must understand that work done is the force applied to move a body through a certain distance.
So;
Work done = Force x distance
Work done = 20 x 90 = 1800J
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)
Answer:
b. 9.5°C
Explanation:
= Mass of ice = 50 g
= Initial temperature of water and Aluminum = 30°C
= Latent heat of fusion = 
= Mass of water = 200 g
= Specific heat of water = 4186 J/kg⋅°C
= Mass of Aluminum = 80 g
= Specific heat of Aluminum = 900 J/kg⋅°C
The equation of the system's heat exchange is given by

The final equilibrium temperature is 9.50022°C