Here, we are required to determine how fast is you drink, sitting in the cup holder, travelling relative to the car.
- The speed of the drink, sitting in the cup holder, relative to the car is; 0m/s
From the laws of relative motion,
- <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)</em>
- <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)</em>
- <em>when object A and Object B are travelling with speed a and b respectively in the same direction, the speed of Object A relative to B is;. (a - b)when object A and Object B are travelling with speed a and b respectively in opposite directions, the speed of Object A relative to B is; (a+b)when object A and Object B are travelling with speed a and b respectively in the same direction, where speed a = speed b, then the speed of object A relative to object B is; zero(0).</em>
Evidently, the scenario in the question is similar to the third scenario above. The cup, sitting in the cup holder is travelling with the car at the same constant speed 10m/s.
Therefore, the speed of the drink relative to the car is zero(0).
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If you stir the juice it increases the surface area.
(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
<u>The given parameters include:</u>
- constant velocity of the elevator, u₁ = 10 m/s
- initial velocity of the ball, u₂ = 20 m/s
- height of the boy above the elevator floor, h₁ = 2 m
- height of the elevator above the ground, h₂ = 28 m
To calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.

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Answer:
Explanation:
1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.
9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2
Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations. We are going to use vf = vi + at. Everything is just given, or we can assume, so I'll just solve.
vf = vi + at
vf = 0 + 127008 km/hour^2 * 24 hours
vf = 3,048,192 km/hour
If there's anything that doesn't make sense let me know.
Answer:
[1, 6, -2]
Explanation:
Given the following :
Initial Position of spaceship : [3 2 4] km
Velocity of spaceship : [-1 2 - 3] km/hr
Location of ship after two hours have passed :
Distance moved by spaceship :
Velocity × time
[-1 2 -3] × 2 = [-2 4 -6]
Location of ship after two hours :
Initial position + distance moved
[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]
= [3-2, 2+4, 4-6] = [1, 6, -2]