Answer:
The answers to the questions have been solved in the attachment.
Explanation:
The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.
Answer:
30 degrees
Explanation:
Reflects off of mirror 1 at 60 degrees....this makes it incident to second mirror at 30 degrees ....then angle of reflection equals this angle of incidence = 30 degrees
See atached diagram
The answer is: Heat can<span> be </span>transmitted<span> though </span>empty space<span> by thermal radiation. Thermal radiation (often called infrared radiation) is a type electromagnetic radiation (or light). Radiation is a form of energy transport consisting of electromagnetic waves traveling at the speed of light.</span>
Answer:
Group and Periods of the Periodic Table of Elements. The three major groups on the Periodic Table are the metals, nonmetals and metalloids. Elements within each group have similar physical and chemical properties.
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
