<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
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Aluminum has an atomic number of 13, so it has 13 protons. In its natural state, it does not have a charge, so it has an equal amount of electrons, 13. The atomic mass of aluminum is approximately 27. Since protons and neutrons make up that mass, and each of the particles are around 1 atomic mass unit, if there are 13 protons, then there are 27-13 neutrons, or 14 neutrons. So the first answer choice is correct.
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
- The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.
Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, <em>C</em> represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;

The steepness of the slope is therefore;

Where;
= Half the width of the truck =
= 1.2 m
= The elevation of the center of gravity above the ground = 2.2 m



The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.
Learn more here:
brainly.com/question/20793607
Answer:
990 J
Explanation:
Kinetic energy is:
KE = ½ mv²
Given m = 55 kg and v = 6 m/s:
KE = ½ (55 kg) (6 m/s)²
KE = 990 J