Knowing the force and distance allows you to calculate the work done.
Then, if you know the TIME, you can calculate the power.
Solve for the linear/tangential speed:
<em>a</em> = <em>v</em>²/<em>r</em>
where <em>a</em> = centripetal acceleration, <em>v</em> = speed, and <em>r</em> = radius.
4.7 m/s² = <em>v</em>²/(0.3 m)
<em>v</em>² = (0.3 m) (4.7 m/s²)
<em>v</em> ≈ 3.96 m/s
For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2<em>π</em> (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is
(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s
Take the reciprocal of this to get the period:
1 / (7.46 rev/s) ≈ 0.134 s/rev
So it takes the record about 0.134 seconds to complete one revolution.
First you find the frequency which is. 1/T(period) so u get 1*10^10
Now that you know frequency u can solve for speed: s= wavelength* frequency, but don't forget to convert 3cm into meters: 3/100= .03m so now you do (1*10^10)(.3)= 300,000,000 m/s or 3*10^8 m/s
Hope this helped :))
Answer:
Part a)
Part b)
Part c)
Since we know that the base area will remain same always
so here the length and width of the object is not necessary to obtain the above data in such type of questions
Explanation:
Part a)
As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force
So here we know
buoyancy force is given as
Now we know that the weight of the cylinder is given as
now we have
Part b)
When the same cylinder is floating in other liquid then we will have
so we have
Part c)
Since we know that the base area will remain same always
so here the length and width of the object is not necessary to obtain the above data in such type of questions
Diagram a shows you a plate tectonic not a boundary and diagram b shows you a convergent plate boundary because the plates are coming together and the oceanic plate is being subducted undneath the continental plate
