If a force of 3000 N is applied to a large rock, but the rock does not move, how much work is done on the rock?
1 answer:
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The period of the orbit would increase as well
Explanation:
We can answer this question by applying Kepler's third law, which states that:
"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"
Mathematically,
Where
T is the orbital period
a is the semi-major axis of the orbit
In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, : but as we saw from the previous equation, the orbital period of the Earth is proportional to
, therefore as
increases, T increases as well.
Therefore, the period of the orbit would increase.
Learn more about Kepler's third law:
#LearnwithBrainly
Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
