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yan [13]
4 years ago
15

We experience fictitious forces due to: a. Rotation of a reference frame b. Inertial reference frames c. Translational motion d.

Universal gravitation.
Physics
1 answer:
lisabon 2012 [21]4 years ago
8 0

Answer:

A.

Explanation:

A fictional force (also called force of inertia, pseudo-force, or force of d'Alembert, 5), is a force that appears when describing a movement with respect to a non-inertial reference system, and that therefore it does not correspond to a genuine force in the context of the description of the movement that Newton's laws are enunciated for inertial reference systems.

The forces of inertia are, therefore, corrective terms to the real forces, which ensure that the formalism of Newton's laws can be applied unchanged to phenomena described with respect to a non-inertial reference system. The correct answer is A.

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A wagon is accelerating down a hill. Which statement is true?
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Potential energy decreases and kinetic energy increases.

Potential energy is related to the height, since the wagon is going downhill, height decreases and potential energy decreases.

Kinetic energy is related to the speed, since the wagon is speeding up, kinetic energy increases.
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To calculate work done on an object, _____. A. multiply the force in the direction of motion by the distance the object moved B.
o-na [289]
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Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

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

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 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

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
3 years ago
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GuDViN [60]

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hope this helps!

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4 years ago
If I accidentally touch a stove and burn myself, which type of heat transfer is occurring?
allochka39001 [22]

Answer:

radiation is the correct answer

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