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Leya [2.2K]
2 years ago
11

ou purchase a rectangular piece of metal that has dimen- sions 5.0 * 15.0 * 30.0 mm and mass 0.0158 kg. The seller tells you tha

t the metal is gold. To check this, you compute the average density of the piece. What value do you get? Were you cheated?
Physics
1 answer:
Natalija [7]2 years ago
4 0

Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;

Density = Mass/Volume

Note that the unit of both mass and volume must be standard unit.

Given mass = 0.0158kg

Dimension of the metal = 5mm×15mm×30mm

Note that 1mm = 0.001m

The volume of the metal will be

0.005×0.015×0.03

= 0.00000225m³

Density = 0.0158/0.00000225

Average density of the metal = 7022.2kg/m³

Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.

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An overhead projector lens is 32.0 cm from a slide (the object) and has a focal length of 30.1 cm. What is the magnification of
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Answer: 15.8

Explanation:

You are given that the

Object distance U = 32 cm

Focal length F = 30.1 cm

First calculate the image distance V by using the formula

1/F = 1/U + 1/V

Substitute F and V into the formula

1/30.1 = 1/32 + 1/V

1/V = 1/30.1 - 1/32

1/V = 0.00197259

Reciprocate both sides

V = 506.94 cm

Magnification M is the ratio of image distance to object distance.

M = V/U

substitute the values of V and U into the formula

M = 506.94/32

M = 15.8

Therefore, the magnification of the image is 15.8 or approximately 16.

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2 years ago
Anne is working on a research project that involves the use of a centrifuge. Her samples must first experience an acceleration o
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Answer:

The final rotation is 2.449 times the initial rotation

Explanation:

Centripetal acceleration and rotation speed of particle is related as

a=rω²    

where ω  is angular speed and r radius of circle

Rearranging equation    

ω=\sqrt{\frac{a}{r} }

The initial acceleration is given by

            a₁ =100g

The final acceleration of particle is

            a₂ = 6 (100g)

                  =600g

The final and initial rotation speed are related as

                  \frac{w1}{w2} =\sqrt{\frac{a2}{a1 }

                  ω2=\sqrt{\frac{600g}{100g} } *w1

                   ω2=2.449ω1

6 0
2 years ago
A student is conducting a physics experiment and rolls four different-
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Answer:

<em>I think the answer is C</em>

<em />

Explanation:

Its heavy but not too heavy

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2 years ago
The diagram shows movement of thermal energy. At bottom a fire has red curved lines labeled Y with arrowheads pointing upward to
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Answer:

X and Z

Explanation:

Conduction occurs through direct physical contact.  Heat transferred from the pot to the handle, and from the handle to the hand, are both examples of conduction.

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Read 2 more answers
A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric
Tcecarenko [31]

Answer:

a)   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m  and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s   c) v_{z}  = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

         Fe + Fm = m a

         a = (Fe + Fm) / m

the electric force is

         Fe = q E   k ^

         Fe = 4.25 10-4 60 k ^

         Fe = 2.55 10-2 k ^

the magnetic force is

         Fm = q v x B

         Fm = 4.25 10⁻⁴  \left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]

         fm = 4.25 10⁻⁴ (-j ^ 30 4)

         fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

      aₓ = 0

there is no force

Axis y

      ay = -5.10 10²/5 10⁻⁵ j ^

      ay = -1.02 107 j ^ / s2

z axis

      az = 2.55 10⁻² / 5 10⁻⁵ k ^

      az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

           x = vo t + ½ aₓ t₂2

           x = 30 0.02 + 0

           x = 0.6m

       

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

           y = I + go t + ½ ay t²

           y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

           y = 1 - 2.04 10³

           y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

          z = zo + v₀ t + ½ az t²

          z = 0 + 0 + ½ 5.1 10² 0.02²

          z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

 since there is no acceleration the speed remains constant

          vₓ = 30.0 m / s

Axis y

  let's use the equation v = v₀ + a_{y} t

         v_{y} = 0 + -1.02 10⁷ 0.02

          v_{y} = 2.04 10⁵ m / s

z axis

          v_{z} = vo + az t

          v_{z} = 0 + 5.1 10² 0.02

          v_{z}  = 1.02 10⁻¹m / s

8 0
2 years ago
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