In this question, you're determining the time (t) taken for an object to fall from a distance (d).
The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)
d = 8,848m and g = 9.8m/s
Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)
The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)
Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)
Now for the last step, find the square root of the remaining number:
t = 42.5s
So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.
I hope this helps :)
Answer:
the radii of curvature is 30 cm.
Explanation:
given,
object is place at = 45 cm
image appears at = 90 cm
focal length = ?
refractive index = 1.5
radii of curvature = ?


f = 30 cm
using lens formula





R = 30 cm
hence, the radii of curvature is 30 cm.
Technically you can go forever on and on, but maybe your question was like how many rotations in a certain distance?
Answer:
Change in temperature ∆(tita) is 266.097°C
Explanation:
Ok kinectic energy = 1/2MV²
5.4 grams =( 5.4/1000) kilogram
Kinectic energy =( 1/2 )*(5.4/1000)*261²
Kinectic energy = 183.9267 joules
If kinetic energy = thermal energy
183.9267 joules = mc∆(tita)
Where ∆(tita) = change in temperature
And c = 128 J/kg
∆(tita) = 183.9267/((5.4/1000)*128)
∆(tita) = 266.097
∆(tita) = 266.097°C
Explanation:
Let us assume that the separation of plate be equal to d and the area of plates is
. As the capacitance of capacitor is given as follows.
C = 
It is known that the dielectric strength of air is as follows.
E = 
Expression for maximum potential difference is that the capacitor can with stand is as follows.
dV = E × d
And, maximum charge that can be placed on the capacitor is as follows.
Q = CV
= 
= 
= 
= 
or, = 10.62 nC
Thus, we can conclude that charge on capacitor is 10.62 nC.