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marta [7]
3 years ago
12

This problem uses the same concepts as Multiple-Concept, except that kinetic, rather than static, friction is involved. A crate

is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal.
The coefficient of kinetic friction between the crate and the ramp is 0.360. Find the acceleration of the moving crate.
Physics
1 answer:
Bingel [31]3 years ago
3 0

Answer:

1.96 m/s^2

Explanation:

angle of inclination, θ = 30.6°

coefficient of friction, μ = 0.36

Acceleration of the crate

a = g Sin θ - μ g Cos θ

a = 9.8 ( Sin 30.6 - 0.36 x Cos 30.6)

a = 9.8 ( 0.509 - 0.309)

a = 1.96 m/s^2

Thus, the acceleration f the crate is 1.96 m/s^2.

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vlada-n [284]

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

Centripetal force - a force acts on an moving object in circular path.

the centripetal force is given by

F= mv²/r       (equation1)

Work done is given by

W = Fd          (equation 2)

d = 2π

work is done by the centripetal force on mass m during an angular displacement of 2π revolutions is given by:

to calculate work done using equation 1 in 2  we get

W = mv² d/r

 W = mv² × 2π /r J

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

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6 0
1 year ago
Which of the following is a combustion reaction?
Nadya [2.5K]

Answer:

Option C

Explanation:

Hydrocarbon + Oxygen = Carbon dioxide + Water

is example of combustion reaction

6 0
2 years ago
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Does heating a cup of water allow it to dissolve more sugar? Temperature of the water is measured in degrees Centigrade. Amount
mixas84 [53]

Yes heating water allows it to dissolve more Sugars because the molecular distance increases and this distance can be covered by more sugar. In the given question, The independent variable would be the temperature of water.

Since to whatever temperature the water boils at the boiling temperature of  does not change remains hundred degree. Rest all the variables can vary the weight of the amount of sugar with the variable in the temperature of Boiling of water to remain constant.

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3 years ago
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

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Also, work= force X distance. The energy does not go away, but rather get changed into some other form of energy
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