Moe is carrying and supporting a large crate by exerting 1000 N. He is walking at constant velocity across a level, horizontal f
1 answer:
To solve this problem it is necessary to apply the agreements related to Strength and Work.
Work is defined as the amount of energy required to move a force over a certain distance.
Where,
F = Force
d = distance
Our values are given as,
F = 1000N
d = 10m
Therefore replacing,
W = 1000*10
W = 10000J
Therefore the neccesary work is 10000J or 10kJ.
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Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.
In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.
Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.
We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.
Answer:
∆h = 0.071 m
Explanation:
I rename angle (θ) = angle(α)
First we are going to write two important equations to solve this problem :
Vy(t) and y(t)
We start by decomposing the speed in the direction ''y''
Vy in this problem will follow this equation =
where g is the gravity acceleration
This is equation (1)
For Y(t) :
We suppose yi = 0
This is equation (2)
We need the time in which Vy = 0 m/s so we use (1)
So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)
2.236 m is the maximum height from the shell (in which Vy=0 m/s)
Let's calculate now the height for t = 0.555 s
The height asked is
∆h = 2.236 m - 2.165 m = 0.071 m
