Solution :
Given
Diameter of the roulette ball = 30 cm
The speed ball spun at the beginning = 150 rpm
The speed of the ball during a period of 5 seconds = 60 rpm
Therefore, change of speed in 5 seconds = 150 - 60
= 90 rpm
Therefore,
90 revolutions in 1 minute
or In 1 minute the ball revolves 90 times
i.e. 1 min = 90 rev
60 sec = 90 rev
1 sec = 90/ 60 rec
5 sec = 
= 75 rev
Therefore, the ball made 75 revolutions during the 5 seconds.
1) <span>A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar eclipse or an -->
Answer: ANULAR ECLIPSE. Since the moon is too far, it will cover only a part of the sun, and only the external ring of the moon will be visible; this is called anular eclipse.
2) </span><span>anyone looking from the night side of earth can, in principle, see a -->
Answer: LUNAR ECLIPSE. If the moon is the right position, and the Earth's shadow covers partially or totally the moon, then a lunar eclipse occurs.
3) </span><span>during some lunar eclipses, the moon's appearance changes only slightly, because it passes only through the part of earth's shadow called the -->
Answer: PENUMBRA.
4) </span><span>a ... can occur only when the moon is new and has an angular size larger than the sun in the sky -->
Answer: TOTAL SOLAR ECLIPSE. When the moon is new, it means it is between the sun and the Earth, and its dark side faces the Earth. If the moon's angular size is also larger than the sun angular size, than it will completely cover the sun, and a total solar eclipse occurs.
5) </span><span>a partial lunar eclipse begins when the moon first touches earth's -->
Answer: SHADOW. The Earth's shadow will start to cover the moon, and partial lunar eclipse will start.
6) </span><span> a point at which the moon crosses earth's orbital plane is called a(n) -->
Answer: NODE. Eclipses occur only when the Moon is at or close to a node, otherwise sun, earth and moon are not "aligned".</span>
The magnetic field strength of a very long current-carrying wire is proportional to the inverse of the distance from the wire. The farther you go from the wire, the weaker the magnetic field becomes.
B ∝ 1/d
B = magnetic field strength, d = distance from wire
Calculate the scaling factor for d required to change B from 25μT to 2.8μT:
2.8μT/25μT = 1/k
k = 8.9
You must go to a distance of 8.9d to observe a magnetic field strength of 2.8μT
Answer:
(A) 140 j/sec (b) 1.26 K
Explanation:
We have given the heat heat flowing into the refrigerator = 40 J/sec
Work done = 40 W
(a) So the heat discharged from the refrigerator 
(b) Total heat absorbed =140 j/sec 
Let the temperature be 
Heat absorbed per hour =504000 ![[tex]=400\times 10^3\times \Delta T](https://tex.z-dn.net/?f=%5Btex%5D%3D400%5Ctimes%2010%5E3%5Ctimes%20%5CDelta%20T)
So 
