Answer:
F = 24 N
Explanation:
In this exercise we have a bar l = 100 m with a center of gravity x = 4 m, which force is needed to lift it from the other end
Let's use the rotational equilibrium relationship, where we consider the counterclockwise rotations as positive and fix the reference system at the point closest to the center of gravity
∑ τ = 0
F l -x W = 0
F = 
let's calculate
F = 
4/100 600
F = 24 N
Answer: C) divide: distance ÷ velocity
Explanation:
The velocity 
equation is distance 
divided by time 
:
If we isolate we will have:
Hence, the correct option is C: distance divided by velocity.
Answer:
300000 J / 400 J/s = 750 s = 12.5 minutes
Explanation:
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t=(0-(250sin75)^2)/-9.8
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>
<span>the max height one is d=0.5*9.8*t^2 </span>
<span>d= max height subtract 1800-d</span>
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
