The charge on the moving particle
Answer:
I took 3*sqrt(10/83)= 1.110349815
And rounded to 1.11 Hz
Explanation:
Answer:
V = 0.0723 volts = 72.3 milivolts
Explanation:
The emf induced in the rod is the motional emf due to the magnetic field. This motional emf can be calculated by the following formula:

where,
V = Motional EMF = ?
v = speed of rod = 12.5 m/s
B = Magnetic Field = 6.23 mT = 0.00623 T
l = Length of rod = 92.9 cm = 0.929 m
θ = angle between v and B = 90°
Therefore,

<u>V = 0.0723 volts = 72.3 milivolts</u>
<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:
Yes
Explanation:
An increased speed will result in an increased amount of energy, so when it crashes some of that energy will bounce back and crumple the car.