Answer:
The magnitude of the angular acceleration is 
Explanation:
From the question we are told that
The angular speed of CD is 
time taken to decelerate is 
The final angular speed is 
The angular acceleration is mathematically represented as
substituting values
The negative sign show that the CD is decelerating but the magnitude is
<span>Δ</span>E = q + w
q = heat (quantity of)
q and w can be positive or negative depending on if work/heat is being absorbed/done on the system or released/done by the system
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
The total power emitted by an object via radiation is:
where:
A is the surface of the object (in our problem,
is the emissivity of the object (in our problem,
)
is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is
Substituting these values, we find the power emitted by radiation:
So, the correct answer is D.
Answer:
31.25 m
25m/sec
Explanation:
Given :-
Time = 5sec
V = 0 (in going up)
U = 0 (in comming down)
Find :-
H and U by which it is thrown up
Since the total time is 5 sec ,therefore half time will be taken to go up and another half will be taken to go down .
We know that ,
V = U + gt
0 = U - 10*2.5
U = 25 m/sec
Also,
V² = U² +2gs
0 = 625 - 20s
s = 625/20 = 31.25 m
