Kinetic energy lost in collision is 10 J.
<u>Explanation:</u>
Given,
Mass,
= 4 kg
Speed,
= 5 m/s
= 1 kg
= 0
Speed after collision = 4 m/s
Kinetic energy lost, K×E = ?
During collision, momentum is conserved.
Before collision, the kinetic energy is

By plugging in the values we get,

K×E = 50 J
Therefore, kinetic energy before collision is 50 J
Kinetic energy after collision:


Since,
Initial Kinetic energy = Final kinetic energy
50 J = 40 J + K×E(lost)
K×E(lost) = 50 J - 40 J
K×E(lost) = 10 J
Therefore, kinetic energy lost in collision is 10 J.
<span>The number in front is the number of molecules (or atoms) taking part in the (balanced) chemical reaction equation.</span>
Answer:
Given:
Fundamental frequency: 470Hz
T1:310k,T2:315k
Calculating velocity
Recall v=(331m/s)✓[T1/273k)
V=331✓(310/273)
V1=331*(1.0656)=352.72m/s
V2=331✓(315/273)=355.5m/s
Fundamental frequency=4L
F2=F1(V2/V1)
F2=470(355.5/352.72)=474.4Hz
Beat=[F2-F1]=474.4-470=4.4Hz
Explanation:
The correct answer to go in the blank would be A) The particles are moving faster.
Answer:
Magnification of the objective lens used and the magnification of the ocular lens.
Explanation: I hope you have/had an amazing day today<3