We are given the equation:
<span>x = 11t^2
</span>
We use that equation to calculate for the distance traveled.
For (a)
At t=2.20 sec,
x =53.24 meters
At t=2.95 sec,
x =95.73 meters
Velocity = (95.73 meters - 53.24<span> meters) / (2.95 s - 2.20 s ) = 56.65 m/s
</span>For (b)
At t=2.20 sec,
x =53.24 meters
At t=2.40 sec,
x =63.36 meters
Velocity = (63.36 meters - 53.24<span> meters) / (2.40 s - 2.20 s ) = 50.6 m/s</span>
<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: a
Explanation: it talks about how they live without that much light
The re<span>sistance of the second wire is 16 R.
where R is the resistance of the first wire.
R = </span>ρ

where l = length of the wire
A = area of the wire
A =

where, r =

Thus, on finding the ratio of resistance of the two wires, we get,

here, R1 = R
l1 = 8m
l2 = 2m
A1=π

A1=π

we get. R2 = 16R