When you push on an object such as a wrench, a steel pry bar, or even the outer edge of a door, you produce a torque equal to th
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The distance covered by the object between t =4 s and t = 6 s is 4 m
Explanation:
In a velocity-time graph, the distance covered by the object represented can be found by calculating the area under the curve.
Therefore, the distance covered by the object between t = 4 s and t = 6 s is the area under the curve between 4 s and 6 s.
We see that we have to calculate the area of a triangle, with:
Base:
And height:
Therefore, the area is
So, the distance covered by the object is 4 m.
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Answer:
a) h = 593.50 m
b) h₁₁ = 103 m
c) vf = 107.91 m/s
Explanation:
a)
We will use second equation of motion to find the height:
where,
h = height = ?
vi = initial speed = 0 m/s
t = time taken = 11 s
g = 9.81 /s²
Therefore,
<u>h = 593.50 m</u>
b)
For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:
where,
vf = final velocity at tenth second = v₁₀ = ?
t = 10 s
vi = 0 m/s
Therefore,
Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:
where,
h = height covered during last second = h₁₁ = ?
vi = v₁₀ = 98.1 m/s
t = 1 s
Therefore,
<u>h₁₁ = 103 m</u>
c)
Now, we use first equation of motion for complete motion:
where,
vf = final velocity at tenth second = ?
t = 11 s
vi = 0 m/s
Therefore,
<u>vf = 107.91 m/s</u>