The slope of a speed-time graph is the acceleration represented by the graph.
All other parts of this question refer to a lab experiment or exercise
where I was not present, but Zeesam16 was. Therefore I have no data
with which to answer the rest of the question, and hope that Zeesam can
handle it.
Answer:
time required is 6.72 years
Explanation:
Given data
mass m = 3.20 ✕ 10^7 kg
height h = 2.00 km = 2 × 10^3 m
power p = 2.96 kW =2.96 × 10^3 J/s
to find out
time period
solution
we know work is mass × gravity force × height
and power is work / time
so we say that power = mass gravity force × height / time
now put all value and find time period
power = mass × gravity force × height / time
2.96 × 10^3 = 3.20 ✕ 10^7 × 9.81× 2 × 10^3 / time
time = 62.784 × 10^10 / 2.96 × 10^3
time = 21.21081081 × 10^7 sec
time = 58.91891892 × 10^3 hours
time = 6.72 years
so time required is 6.72 years
The mass of the object will remain the same rather it's on the moon or on the Earth and even in other places. But the weight will change on the moon, so its weight will be different from the one it had on Earth
To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.
We know that by Hooke's law
Where,
k = Spring constant
x = Displacement
Re-arrange to find k,
Perioricity in an elastic body is defined by
Where,
m = Mass
k = Spring constant
Therefore the period of the oscillations is 0.685s
