K.E=1/2mv^2 K.E=1/2multiply1multiply8^2=32joules
1.A) 4.9 m
AL2006 Ace
The instant it was dropped, the ball had zero speed.
After falling for 1 second, its speed was 9.8 m/s straight down (gravity).
Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.
Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.
ANYTHING you drop does that, if air resistance doesn't hold it back.
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2 idk sorry
<span>You can start with the equations you know
a=v^2/r = (2pi*r/T)^2/r = 4pi^2r/T^2
Radius of earth (R) = 6378.1 km
Time in one day (T) = 86400 seconds
Latitude = 44.4 degrees
If you draw a circle and have the radius going out at a 44.4 degree angle above the center you can then find the r.
r=Rcos(44.4)
r=6378.1cos(44.4)
r= 4556.978198 km or 4556978 m
Now you can plug this value into the acceleration equation from above...
a= 1.8*10^8/7.47*10^9
a= .0241 m/s^2 </span>
Answer:
it would be 3
Explanation:
because you have to divide the length by the height of the incline.
Answer:
Explanation:
Area A of the coil = .1 x .1 = .01 m²
no of turns n = 5
magnetic field B = .5 t²
Flux Φ perpendicular to plane passing through it.= nBA sin30
rate of change of flux
dΦ/dt = nAdBsin30 / dt
= nA d/dt (.5t²x .5 )
= nA x 2 x .25 x t
At t = 4s
dΦ/dt = nA x 2
= 5x .01 x 2
= .1
current = induced emf / resistance
= .1 / 4
= .025 A
= 25 mA.
