The total displacement of the toy car at the given positions is 0.
The given parameters;
- <em>First displacement of the car, = 5 cm left</em>
- <em>Second displacement of the car, = 8 cm right</em>
- <em>Third displacement of the car, = 3 cm to the left</em>
The total displacement of the car is calculated as follows;
- <em>Let the </em><em>left </em><em>direction be "</em><em>negative </em><em>direction"</em>
- <em>Let the </em><em>right </em><em>direction be "</em><em>positive </em><em>direction"</em>
Thus, the total displacement of the toy car at the given positions is 0.
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Answer:
2156J
Explanation:
Given parameters:
Height of lift = 10m
Mass = 22kg
Unknown:
Work done by the machine = ?
Solution:
Work done is the force applied to move a body through a certain distance.
So;
Work done = Force x distance
Here;
Work done = mass x acceleration due to gravity x height
Work done = 22 x 9.8 x 10 = 2156J
Answer:
Explanation:
The Newton's law in this case is:
Here, is the air temperture, C and k are constants.
We have
in
So:
And we have in , So:
Now, we have:
Applying (1) for :
Applying (1) for :
Answer:
4.2 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v
315 g m/s − 525 g m/s = -525 g m/s + (75 g) v
315 g m/s = (75 g) v
v = 4.2 m/s
The letter i is used to signify that a number is an imaginary number. It stand for the square root of negative one.