Answer:
(a) 89 m/s
(b) 11000 N
Explanation:
Note that answers are given to 2 significant figures which is what we have in the values in the question.
(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

(b) The tension,
, is given by

where
is the speed,
is the tension and
is the mass per unit length.
Hence,

To determine
, we need to know the mass of the cable. We use the density formula:

where
is the mass and
is the volume.

If the length is denoted by
, then


The density of steel = 8050 kg/m3
The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is




Answer:
Time period of the osculation will be 2.1371 sec
Explanation:
We have given mass m = (B+25)
And the spring is stretched by (8.5 A )
Here A = 13 and B = 427
So mass m = 427+25 = 452 gram = 0.452 kg
Spring stretched x= 8.5×13 = 110.5 cm
As there is additional streching of spring by 3 cm
So new x = 110.5+3 = 113.5 = 1.135 m
Now we know that force is given by F = mg
And we also know that F = Kx
So 

Now we know that 
So 


Answer:
Hi myself Shrushtee.
Explanation:
Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artificial gravity, human growth is stunted and biological functions break down. An effective way to create artificial gravity is through the use of a rotating enclosed cylinder, as shown in the figure. Humans walk on the inside edge of the cylinder, which is sufficiently large (diameter of 2235 meters) that its curvature is not readably noticeable to the inhabitants. (The space station in the figure is not drawn to the scale of the human.) Once the space station is rotating at the necessary speed, how many minutes would it take the space station to make one revolution?
The distance traveled by the man in one revolution is simply the circumference of the space station, C = 2p R. From this result, you should be able to deduce the time it takes for the space station to sweep out a complete revolution.
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