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Serggg [28]
4 years ago
7

A skateboarder wants to cross a large playground and notices that there are large shapes painted on its asphalt surface. One sha

pe is a large circle that covers the entire playground. Inscribed in that circle is a square, with one point at the skateboarder’s feet and the other at his destination. The surface is rough and, therefore, friction is present. What path does he follow in order to minimize the work done by nonconservative forces?
Physics
1 answer:
Elden [556K]4 years ago
6 0
<h2>Diagonal of circle </h2>

Explanation:

As the skateboarder wants to cross the play ground . The surface is rough .

As we know , the force of friction is non-conservative force . Thus work is required against this force .

We have formula:

work done = Force x distance (in one direction )

Te force applied cannot be changed , so he is to decrease the distance .

In case of circle , diameter is the minimum distance . Thus he is supposed to move along it .

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A centrifuge in a medical laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates through 52.0 re
eduard

The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

<h3> Constant angular acceleration</h3>

Apply the following kinematic equation;

ωf² = ωi² - 2αθ

where;

  • ωf is the final angular velocity when the centrifuge stops = 0
  • ωi is the initial angular velocity
  • θ is angular displacement
  • α is angular acceleration

ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s

θ = 52 rev x 2π rad/rev = 326.7 rad

0 = ωi² - 2αθ

α = ωi²/2θ

α = ( 356.05²) / (2 x 326.7)

α = 194.02 rad/s²

Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

Learn more about angular acceleration here: brainly.com/question/25129606

#SPJ1

7 0
2 years ago
A force of 8800 N is exerted on a piston that has an area of 0.010 m^2. What is the force exerted by a second piston that has an
professor190 [17]

Explanation:

It is given that,

Force on piston, F₁ = 8800 N

Area, A_1=0.01\ m^2

Area, A_2=0.04\ m^2

Let F₂ is the force exerted on the second piston. Using Pascal's law as :

Pressure at piston 1 = Pressure at piston 2

\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}

F_2=\dfrac{F_1}{A_1}\times A_2

F_2=\dfrac{8800}{0.01}\times 0.04

F_2=35200\ N

So, the force exerted by a second piston is 35200 N. Hence, this is the required solution.

8 0
3 years ago
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A 3.2 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4 + 15.7 x − 1.5 x 2 , where Fx
garik1379 [7]

Answer:

Explanation:

Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as

W = F×d

or

W = \int\limits^b_a {Fx} \, dx.................................. Equation 1

Where b = upper limit, a = lower limit, Fx = expression of force.

<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>

Substituting these values into equation 1

<em>W = \int\limits^a_b {(4 + 15.7x - 1.5x^{2} )dx} \,</em>

W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ

Work = upper limit - lower limit

Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2

Substituting the values of a and b into equation 2

Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]

Work = [5.2 + 26.53 -3.29 + C] - C

Work = 28.44 J

Work done by the force = 28.44 J.

8 0
3 years ago
Reference frame definitely changes when what also changes?
kompoz [17]

Explanation is in the file

tinyurl.com/wpazsebu

8 0
3 years ago
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Which of the following statements is true?
erastova [34]

Hi there!

Question - Which of the following statements is true?

Answer - C. You are exposed to nuclear radiation every day.

Why - "radiation in the form of elementary particles emitted by an atomic nucleus, as alpha rays or gamma rays, produced by decay of radioactive substances or by nuclear fission."

6 0
3 years ago
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