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Solnce55 [7]
3 years ago
5

Which of the following statements is/are true? Check all that apply. A. A potential energy function can be specified for a conse

rvative force. B. A nonconservative force permits a two-way conversion between kinetic and potential energies. C. The work done by a nonconservative force depends on the path taken. D. A potential energy function can be specified for a nonconservative force. E. A conservative force permits a two-way conversion between kinetic and potential energies. F. The work done by a conservative force depends on the path taken.
Physics
1 answer:
Otrada [13]3 years ago
5 0

Answer:

A. A potential energy function can be specified for a conservative force.

TRUE

because there is no loss of energy in conservative type of forces.

B. A nonconservative force permits a two-way conversion between kinetic and potential energies.

FALSE

It is not true because energy is not conserved in non-conservative forces.

C. The work done by a nonconservative force depends on the path taken.

TRUE

It depends on total path length while in conservative it only depends on initial and final state

D. A potential energy function can be specified for a nonconservative force.

FALSE

Since energy is not stored in non-conservative forces so it is not defined for non conservative forces

E. A conservative force permits a two-way conversion between kinetic and potential energies.

TRUE

Work done against conservative forces is stored in form of potential energy so it is possible to have two way conversion.

F. The work done by a conservative force depends on the path taken.

FALSE

Conservative force work depends only on initial and final state

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Answer:

Negative acceleration occurs when the acceleration vector points to the left.

1. Object slowing down in the positive direction.

2. Object speeding up in the  negative direction.

Following six statements:

1. T

2. F

3. T

4. T

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3 years ago
Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will
zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

     Initial velocity, u  = u sin θ

     Final velocity, v =  0 m/s    

     Acceleration = -g

     Substituting

                      v = u + at  

                      0 = u sin θ - g t

                      t=\frac{usin\theta }{g}

This is the time of flight.

Consider the horizontal motion of ball,

        Initial velocity, u =  u cos θ

        Acceleration, a =0 m/s²  

        Time, t=\frac{usin\theta }{g}  

     Substituting

                      s = ut + 0.5 at²

                      s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}

This is the range.

In this problem

              u = 30 m/s

              g = 9.81 m/s²

              θ = 45° - For maximum range

Substituting

               s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

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Mass Center Determine the coordinates (x, y) of the center of mass of the area in blue in the figure below. Answers: x=(3)/(8)a
Naya [18.7K]

Explanation:

The x and y coordinates of the center of mass are:

xcm = ∫ x dm / m = ∫ x ρ dA / ∫ ρ dA

ycm = ∫ y dm / m = ∫ y ρ dA / ∫ ρ dA

Assuming uniform density, the center of mass is also the center of area.

xcm = ∫ x dA / ∫ dA = ∫ x y dx / A

ycm = ∫ y dA / ∫ dA = ∫ ½ y² dx / A

First, let's find the area:

A = ∫ y dx

A = ∫₀ᵃ (-h/a² x² + h) dx

A = -⅓ h/a² x³ + hx |₀ᵃ

A = -⅓ h/a² (a)³ + h(a)

A = ⅔ ha

Now, let's find the x coordinate of the center of mass:

xcm = ∫ x y dx / A

xcm = ∫₀ᵃ x (-h/a² x² + h) dx / (⅔ ha)

xcm = ∫₀ᵃ (-h/a² x³ + hx) dx / (⅔ ha)

xcm = (-¼ h/a² x⁴ + ½ hx²) |₀ᵃ / (⅔ ha)

xcm = (-¼ h/a² (a)⁴ + ½ h(a)²) / (⅔ ha)

xcm = (¼ ha²) / (⅔ ha)

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Next, we find the y coordinate of the center of mass:

ycm = ∫ y² dx / A

ycm = ∫₀ᵃ ½ (-h/a² x² + h)² dx / (⅔ ha)

ycm = ∫₀ᵃ ½ (h²/a⁴ x⁴ − 2h²/a² x² + h²) dx / (⅔ ha)

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ycm = ½ (⅕ h²/a⁴ (a)⁵ − ⅔ h²/a² (a)³ + h² (a)) / (⅔ ha)

ycm = ½ (⁸/₁₅ h²a) / (⅔ ha)

ycm = ⅖ h

6 0
3 years ago
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