Answer:
1. It undergoes reflection. 2. It undergoes refraction. 3. It undergoes diffraction.
Explanation:
1. It undergoes reflection. This is because it bounces off surfaces when incident on them.
2. It undergoes refraction. This is because it changes direction when it passes from one medium to another
3. It undergoes diffraction. This is because it spreads out when it passes through doors and windows similar in dimension to the dimensions of its wavelength
from rarefaction to rarefaction for a longitudinal wave
The driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.
The given parameters:
- <em>Speed of the driver observed by the stationary police officer, Vo = 44.7 m/s</em>
- <em>Speed of the driver, V = 26.8 m/s.</em>
- <em>Speed limit = 60 mph</em>
The speed limit of the driver in meter per second is calculated as follows;
From the speed limit, it is obvious that the driver's speed is within the limit. Thus, we can conclude that the driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.
Learn more about relative velocity here: brainly.com/question/17228388
Answer: hope it helps you...❤❤❤❤
Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.
No, not necessarily.
For instance, Newton’s 2nd law is F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.
But take a look at this (incorrect) equation for the force of gravity:
F=−G(m+M)Mm√|r|3r
It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.
It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.
A simpler counter example is 1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.
Answer:
The value of mass 1, m1= 6/5m
The value of mass 2, m2= 3/5m
Explanation:
case 1:
here tension and the acceleration will be:
for m1;
- mg-T=ma
- 2mg - 2T = 2ma .....1.
for m2:
adding the both equations,
2mg - 2T + 2T-mg = 2ma + ma/2
a = 2/5 g
putting the value of a into the equation 1.
mg - T = m* (2/5)g
T = 3/5 ( mg )
now
case 2:
The two identical blocks are released from the rest, the tension remains the same as the case 1.
so,
for m1:
for m2:
adding both equations we get,
2T-m2g + 2m2g - 2T = 0
m2 = m1 / 2
T = m1*g / 2
here we know that
T (case1) = T (case2)
3/5 ( mg ) = m1*g / 2
m1 = 6/5 m
hence
m2 = 3/5 m
learn more about tension here:
<u>brainly.com/question/23590078</u>
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