Does the rankine degree represent a larger or smaller temperature unit than the kelvin degree
1 answer:
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Correct question:
Consider the motion of a 4.00-kg particle that moves with potential energy given by
a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?
b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?
Answer:
a) 3.33 m/s
b) 0.016 N
Explanation:
a) given:
V = 3.00 m/s
x1 = 1.00 m
x = 5.00
At x = 1.00 m
= 4J
Kinetic energy = (1/2)mv²
= 18J
Total energy will be =
4J + 18J = 22J
At x = 5
= -0.24J
Kinetic energy =
= 2Vf²
Total energy =
2Vf² - 0.024
Using conservation of energy,
Initial total energy = final total energy
22 = 2Vf² - 0.24
Vf² = (22+0.24) / 2
= 3.33 m/s
b) magnitude of force when x = 5.0m
At x = 5.0 m
= 0.016N
Answer:
162 km
Explanation:
A diagram can be helpful.
Using the law of cosines, we can find the magnitude of the distance (c) to satisfy ...
c^2 = a^2 +b^2 -2ab·cos(C)
where C is the internal angle of the triangle of vectors and resultant. Its value is ...
180° -39.8° -59.9° = 80.3°
Filling in a=76 and b=156, we get ...
c^2 = 76^2 +156^2 -2·76·156·cos(80.3°) ≈ 26116.78
c ≈ √26116.78 ≈ 161.607
The magnitude of the total displacement is about 162 km.
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Please note that in the attached diagram North is to the right and East is up. That alteration of directions does not change the angles or the magnitude of the result.
