In this scenario, the principles of momentum conservation can be applied since there are no external forces acting on the system. Consequently, the conservation of momentum principle is applicable here. After the bird lands on it, both the bird and the bark will have a unified final speed. Thus, this final speed will be 1 m/s.
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<span>A force of 110 N is applied at an angle of 30</span>°<span> to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.
For the component parallel to the ground:
x = rcos</span>β
<span>x = 110cos30</span>°
<span>x = 95.26
For the component perpendicular to the ground:
y = rsin</span>β
<span>y = 110sin30</span>°
<span>y = 55</span>
The radius of the moon's orbit is calculated as R = 7.715 x 10⁷ m, and the moon's orbital period is T = 14.48 hr. The given orbital speed of the moon is v = 9.3 x 10³ m/s, with Neptune's mass being M = 1.0 x 10²⁶ Kg. The moon's orbital velocity can be expressed using the formula. Therefore, by squaring the equation and resolving for r + h, we calculate: R = GM / v². Upon substituting in, we find R to be 7.715 x 10⁷ m. The relation for the moon's orbital period yields T = 2π/ω and simplistically, T = 2πR/v, where ω = v/r. Following this, we compute T, leading to the conclusion: T = 14.48 hr.
Answer:
The kinetic energy is higher for the first cart.
Explanation:
For the second cart, its mass is 2kg and the momentum measured is 10kg m/s, which leads to
resulting in
.
Consequently, the kinetic energy for the 3kg cart ends up as
indicating it is less than that of the 1kg cart so it follows that the first cart possesses greater kinetic energy.
