Answer is 6 tires.
This is a projectile question.
First make sure units are consistent - express speed in m/s.
20 km/h = 20000m / 3600 s = 5.56 m/s
Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.
Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:
Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s
An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:
Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s
Total trip time:
0.19 x 2 = 0.38s
Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:
v = d / t
5.22 = d / 0.38
d = 1.98m
Now divide this total distance by the length of an individual tire to find the number of tires he will clear:
1.98 / 0.3 = 6.6 tires
Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).
Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!
Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J
Answer:
Explanation:
There are three basic ways to increase the likelihood of safely dropping an egg:
Slow down the descent speed.
Parachutes are an obvious method for slowing the decent speed, as long as the design includes a way to keep the parachute open.
Cushion the egg so that something other than the egg itself absorbs the impact of landing.
The largest end of the egg has an area of air trapped between the egg's two membranes. This air space forms when the contents of the egg cool and contract after the egg is laid. It accounts for the crater you often see at the end of a hard-cooked egg. Upon impact the heavier spherical yolk continues moving towards the ground. The compression of the airspace acts like an air bag for the eggs' valuable contents. Building an artificial cushioning device will also help absorb the impact of landing.
The largest end of the egg has an area of air trapped between the egg's two membranes. This air space forms when the contents of the egg cool and contract after the egg is laid. It accounts for the crater you often see at the end of a hard-cooked egg. Upon impact the heavier spherical yolk continues moving towards the ground. The compression of the airspace acts like an air bag for the eggs' valuable contents. Building an artificial cushioning device will also help absorb the impact of landing.
Orient the egg so that it lands on the strongest part of the shell.
The arch structure at either end of the egg is stronger than its sides. Pressure is distributed down (or up) the arches so that less pressure acts on any one point. Orienting the arch downwards will increase the egg's survival.
Hope this helps you
B
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<span>A- 2*[(m1 - m2)/(m1 + m2)]*g/L
The rotation is in the counterclockwise direction and the angular acceleration is positive.
B- 2*[(m1 - m2)/(m1+ m2 +mbar/3)]*g/L
The rotation is in the counterclockwise direction and the angular acceleration is positive.</span>
