Answer:
0.62 m/s
Explanation:
Given that a model train, travelling at speed , approaching a buffer model train buffer spring
The train, of mass 2.5 kg, is stopped by compressing a spring in the buffer.
After the train has stopped, the energy stored in the spring is 0.48 J.
Calculate the initial speed v of the train
Solution
The total energy stored in the spring will be equal to the kinetic energy of the train.
That is,
1/2fe = 1/2mv^2
Substitutes the spring energy, and mass into the formula
0.48 = 1/2 × 2.5 × V^2
2.5V^2 = 0.96
V^2 = 0.96 / 2.5
V^2 = 0.384
V = sqrt ( 0.384 )
V = 0.62 m/s
Therefore, the initial velocity of the train is 0.62 metres per second.
Answer:
675J
Explanation:
Given parameters:
Force = 45N
Distance = 15m
Unknown:
Work done by Sheila = ?
Solution:
Work done by a body is the amount of force applied to make a body move through a distance;
Work done = Force x distance
Now;
Work done = 45 x 15 = 675J
Answer:
180 m
Explanation:
Case 1.
U = 40 km/h = 11.1 m/s, V = 0, s = 20 m
Let a be the acceleration.
Use third equation of motion
V^2 = u^2 + 2 as
0 = 11.1 × 11.1 - 2 × a × 20
a = 3.08 m/s^2
Case 2.
U = 220 km/h = 33.3 m/s, V = 0
a = 3.08 m/s^2
Let the stopping distance be x.
Again use third equation of motion
0 = 33.3 × 33.3 - 2 × 3.08 × x
X = 180 m
Answer: 6.022×10²³ molecules of CH4 which consists of 1 mole of C atoms and 4 moles of H atoms.
This is done by nuclear fusion in which it converts hydrogen atoms into helium. so the byproduct of nuclear fusion is that in the sun's core is a massive volume of energy that getting release and radiates outward towards of the surface of the sun and into space to the solar systems and a bit beyond. that's kinda all I know hope it helps. =^-^=