The student who did the most work is student 2 with 2500 Joules.
<u>Given the following data:</u>
To determine which of the students did the most work:
Mathematically, the work done by an object is given by the formula;
<u>For </u><u>student 1</u><u>:</u>
Work done = 600 Joules
<u>For </u><u>student 2</u><u>:</u>
Work done = 2500 Joules.
Therefore, the student who did the most work is student 2 with 2500 Joules.
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It’s either C or D. let me know but I would do C!
Refer to the diagram shown below.
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
Answer:
According to Hook's law, we know,
strain/stress =Constant
Explanation: So, the ratio between stress and strain is always constant.
So, if stress is increased, then strain changes in that way so that this ratio always remains constant.
