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BlackZzzverrR [31]
2 years ago
6

A convex mirror has a focal length of -13.0 cm. When you place a 6.00 cm tall pencil 60.0 cm in front of that mirror, what is th

e location of the pencil's image?
Physics
1 answer:
wlad13 [49]2 years ago
6 0

The location of the pencil's image is =16.6cm

<h3>Calculation of image location</h3>

The focal length of the convex mirror (f)= -13.0 cm

The object distance (u) = 60cm

The image distance (v) = xcm

Using the formula,1/f= 1/v + 1/u

Make 1/v the subject of formula,

1/v = 1/f -1/u

1/v = 1/13 - 1/60

1/v = 60-13/780

1/v = 47/780

V = 780/47

V = 16.6 cm

Therefore, he location of the pencil's image is = 16.6cm

Learn more about mirror here:

brainly.com/question/13164847

#SPJ1

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To give solution to the exercise we must use the concepts of Torque, Vector magnitude and vector direction of the forces.

For the given problem we have to

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The net torque acting on the particle is

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\tau_{net} = \sqrt{(8)^2+(-8.9)^2}

\tau_{net} = 11.967Nm

PART B) The direction of the torque is given by,

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\theta = -48.04\°

Therefore the torque direction is 48.04° below the x axis.

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Explanation:

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During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.
Phoenix [80]

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

Q = \frac{\Delta P \pi r^4}{8\eta l}

Where:

\eta_i = are the viscosities of the concrete before and after the increase

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r_1, R_2 = Radio of the vessel before and after the increase

\Delta P= Change in the pressure

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Our values are given as:

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\Delta P_2 = 1.5\Delta P_1 Increase of 50%

Plugging known information to get

Q_1 = \frac{\Delta P \pi r^4}{8\eta l}

Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4

r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}

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Question :

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Hence, this is the required solution.

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