Ok so this is simple projectile motion problem.
if we have an object falling in free fall it is subject to gravity of -9.80m/s^2
so it says it takes 6 sec to fall and we know initial velocity was zero so we know that h=vt+1/2gt^2 so we get h=0+1/2*9.80*6^2 = 176.4m
so solving for final speed we get KE=PE = 1/2mv^2=mgh = 1/2v^2=gh so
v=sqrt(2*g*h) = sqrt(2*9.8*176.4m) = 58.8m/s final speed when it hits the ground
hope this helps you! Thanks!!
By definition of average acceleration,
<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²
Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:
∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N
If you just want the magnitude, drop the negative sign.
Answer:
R = 2216m and The normal force of the seat on the pilot is 5008N
Explanation:
See attachment below please.
Answer:
Average speed = distance/time
From 1 to 9 seconds:
Distance covered = 1 - 0.2 = 0.8 km
Time = 9 - 1 = 8 sec
Average speed = 0.8 km / 8 sec
Average speed = 0.1 km/s .
The average speed for the whole test is 1.6 km / 20 sec = 0.08 km/sec. A graph of speed vs time would average out as a horizontal line at 0.08 km/sec from 1 sec to 21 sec. The area under it would be (0.08 km/s) x (20 sec) = 1.6 km.
Surprise surprise ! The area under a speed/time graph is the distance covered during that time !
In closing, I want to express my gratitude for the gracious bounty of 3 points with which I have been showered. Moreover, the green breadcrust and tepid cloudy water have also been refreshing.
Explanation:
Answer:
Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc) items 3 and 4 are eliminated.
Again item 2 refers to bonds between atoms and is eliminated.
This leaves only item 1.
Nuclear forces are very short range forces between components of the nucleus.
Weak nuclear forces are trillions of times smaller than strong forces.
Gravitational forces are much much smaller than the weak nuclear force.
