Answer:
Part a)
Part b)
Part c)
Part d)
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Explanation:
Part a)
When elevator is ascending with constant speed then we will have
So it will read same as that of the mass
Part b)
When elevator is decending with constant speed then we will have
So it will read same as that of the mass
Part c)
When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have
Reading is given as
Part d)
Here the speed of the elevator is constant initially
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Answer:D.Refractive Indez
Explanation:
It is usually expressed the other way: the ratio of the speed of light in a vacuum to the speed of light in a medium. In that case, it is called the "index of refraction".
I beleive she isnt doing any work due to holding the box motionless, you must be exerting a force in the direction of the box motion. If she is just standing there holding the box their isn't no work becuase no distance has been covered. work = force = distance.
Answer:
The constriction causes the mercury column to break under tension, leaving a vacuum between the bottom of the column and that in the bulb, and the top of the column stays still at the position reached in the body - a "peak hold" system.
Answer:
The width of the central bright fringe on the screen is observed to be unchanged is
Explanation:
To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which
Where,
w = width
wavelength
m is an integer, m = 1, 2, 3...
We here know that as as w are constant, then
We need to find , then
Replacing with our values:
Therefore the width of the central bright fringe on the screen is observed to be unchanged is