16) d = 1/2gt^2
2 m = 1/2(1.62m/s^2)t^2
2 m = 0.81m/s^2*t^2
2.47 s^2 = t^2
t = sqrt(2.47 s^2) = 1.57 s = 1.6 s.
I believe this is the correct way of solving it...
For 22) Around 1000 Km is the best answer since 30 km is too small of an atmosphere...planes fly at 30,000 feet and that is above the troposphere and then we have to go to the stratosphere, mesosphere, thermosphere, and exosphere! 500 m is too small, too, only 0.5 km which does not even pass the troposphere! The ozone layer is at about 100 km, too, and still have the mesophere, thermosphere, etc. to go! 1 million meters is the same as 1000 Km. So, those two!
16) d = 1/2gt^2
2 m = 1/2(1.62m/s^2)t^2
2 m = 0.81m/s^2*t^2
2.47 s^2 = t^2
t = sqrt(2.47 s^2) = 1.57 s = 1.6 s.
I believe this is the correct way of solving it...
15) Some of it is reflected so it hits the bird. The rest is refracted (Snell's law) and hits the fish. So, A.
for 22 ... the answer in the model answer is 30 km !!
maybe they mean in question the distance from the eart's surface to the atmosphere
For 2) conservation of momentum says m1*v1 = (m1+m2)*v2where m1 is mass of woman and m2 is mass of boat. From this we get v2 = v1*m1/(m1+m2), so v2 is smaller than v1. It is in the same direction.
For 16) when you calculate v = x/t you are calculating the average velocity, but you need the final velocity and the initial velocity as the acceleration is given by (final velocity - initial velocity)/t.
The initial velocity is easy enough - it is zero.
average velocity = (initial velocity + final velocity)/2 so: x/t = (0 + final velocity)/2
Rearrange to get: final velocity = 2*x/t.
So acceleration = (final velocity - initial velocity)/t
or 1.62 = (2*2/t - 0)/t
1.62 = 4/t2
t = √(4/1.62)
t = 1.57 s ≈ 1.6 s