<span>What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities
sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s
cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s
Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s.
To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So
0.50 s * 14.7447368 m/s = 7.372368399 m
1.00 s * 14.7447368 m/s = 14.7447368 m
1.50 s * 14.7447368 m/s = 22.1171052 m
2.00 s * 14.7447368 m/s = 29.48947359 m
Rounding the results to 1 decimal place gives
0.50 s = 7.4 m
1.00 s = 14.7 m
1.50 s = 22.1 m
2.00 s = 29.5 m</span>
Answer:
b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface
Explanation:
Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.
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Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
