3 x 108 is roughly 300 people.
Half of them have cars. Half of 300 = 150.
150 x 12000 = 1,800,000 miles driven.
Each car gets 20mpg.
Solve for the # of gallons consumed.
You didn't say so, but we must assume that the "200 km/hr" is
the glider's air-speed, that is, speed relative to the air.
If the air itself is moving at 30 km/hr relative to the ground and
across the glider's direction, then the glider's speed relative to
the ground is
√(200² + 30²)
= √(40,000 + 900)
= √(40,900) = 202.24... km/hr (rounded)
Answer:
you have no picture
Explanation:
if it shows the arrows going towards eachother it is north and south. if it is not it will be either north and north or south and south
Answer:
The speed in the first point is: 4.98m/s
The acceleration is: 1.67m/s^2
The prior distance from the first point is: 7.42m
Explanation:
For part a and b:
We have a system with two equations and two variables.
We have these data:
X = distance = 60m
t = time = 6.0s
Sf = Final speed = 15m/s
And We need to find:
So = Inicial speed
a = aceleration
We are going to use these equation:


We are going to put our data:


With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.



![[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}](https://tex.z-dn.net/?f=%5B%5Csqrt%7B%2815m%2Fs%29%5E2-%282%2Aa%2A60m%29%7D%5D%5E%7B2%7D%3D%5B15m%2Fs-%28a%2A6s%29%5D%5E%7B2%7D)








If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2
After, we are going to determine the speed in the first point:




For part c:
We are going to use:




You first us 1/2(mv^2) to solve for the potential energy and then put that in to PE=m*g*h and solve for hight