The change in distance is 30 because if you subtract both number you'll get 30
20 ohms in parallel with 16 ohm= 8.89
20x16/20+16. Product over sum
Answer:
The average speed of the elevator going down in the abandoned mine is 17.722mph.
Explanation:
If the elevator takes 90 seconds to descend a height of 713m, the average speed of the elevator is:
![v_{av}=x_T/t_T=713m/90s=7.922m/s](https://tex.z-dn.net/?f=v_%7Bav%7D%3Dx_T%2Ft_T%3D713m%2F90s%3D7.922m%2Fs)
And if 1m/s is 2.23694mph, the average speed is:
.
<span>The diver is heading downwards at 12 m/s
Ignoring air resistance, the formula for the distance under constant acceleration is
d = VT - 0.5AT^2
where
V = initial velocity
T = time
A = acceleration (9.8 m/s^2 on Earth)
In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m)
So let's substitute the known values and solve for T
d = VT - 0.5AT^2
-7 = 2.5T - 0.5*9.8T^2
-7 = 2.5T - 4.9T^2
0 = 2.5T - 4.9T^2 + 7
We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164.
Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So
V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s
V = 2.5 m/s - 14.47706141 m/s
V = -11.97706141 m/s
So the diver is going down at a velocity of 11.98 m/s
Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point.
V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s)
V = 2.5 m/s - (-9.477061409 m/s)
V = 2.5 m/s + 9.477061409 m/s
V = 11.97706141 m/s
And you get the exact same velocity, except it's the opposite sign.
In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
Answer:![F_2=16\times F_1](https://tex.z-dn.net/?f=F_2%3D16%5Ctimes%20F_1)
Explanation:
Given
Force of repulsion between two charge particle is given by force F
Electrostatic force is given by
![F=\frac{kq_1q_2}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7Bkq_1q_2%7D%7Br%5E2%7D)
where
and
is the charges of particle
r=distance between charge particle
when charges are doubled and distance is reduced to half
i.e. q become 2 q and r becomes 0.5 r
![F_2=\frac{k(2q)^2}{(0.5r)^2}](https://tex.z-dn.net/?f=F_2%3D%5Cfrac%7Bk%282q%29%5E2%7D%7B%280.5r%29%5E2%7D)
![F_2=\frac{kq^2}{r^2}\times 4\times 4](https://tex.z-dn.net/?f=F_2%3D%5Cfrac%7Bkq%5E2%7D%7Br%5E2%7D%5Ctimes%204%5Ctimes%204)
![F_2=16\times F_1](https://tex.z-dn.net/?f=F_2%3D16%5Ctimes%20F_1)