Answer:
The work done on the suitcase is, W = 600 J
Explanation:
Given,
The average force exerted by Jose on his suitcase, F = 60 N
Jose carried the suitcase to a distance, S = 10 m
The work done on the suitcase is given by the relation
<em>W = F x S</em>
Substituting the given values in the above equation,
W = 60 N x 10 m
= 600 J
Hence, the work done on the suitcase is, W = 600 J
Answer:
a.) Maximum height of the projectile. =32.5m
b.) The time it takes to reach the maximum height=2.6s
c.) Time of flight=2.6s
d.) Range=99.5m
Explanation:
Answer:
t = 2s
Explanation:
When you're looking for instantaneous portions of a graph, of any sort really, it means you're observing a rate at a single point in time [or possibly some other variable]. It's sorta like a snapshot of a rate as opposed to an average rate over an interval. After choosing this rate we'll typically draw a straight, tangent line through it to indicate it's slope. (Tangent lines are just lines that only touch a single point on a graph or shape.)
Another thing to take note of are the values of the graph's major axes. The "y-axis" corresponds to velocity in meters per second, while the "x-axis" corresponds to time in seconds. Normally when relating the two we put "y" over the "x" and say that at any point there are "y[units]" per "x[units]". Though with instantaneous rates, we say the value of "x" is "1"; for reasons I can try to further explain later if you'd like.
With the above information in mind we can turn our attention to your graph. You're told to find the point on this graph where the instantaneous rate of acceleration is -2 m/s². The only place where the graph reflects an instantaneous rate of -2m/s² is at t = 2s. At t = 2, the rate comes out to (2[m/s]/1s), which simplifies to 2m/s². If you then draw the tangent line through the point, you'll find that the line is decreasing (going down from left to right) which means that the instantaneous rate is negative.
So at t = 2s, we have an instantaneous acceleration of -2m/s².
