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Mashutka [201]
2 years ago
12

8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.

Physics
1 answer:
mamaluj [8]2 years ago
5 0

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle (p), measured in centimeters, is represented by the following formula:

p = 2\cdot (w+l) (1)

Where:

w - Width, measured in centimeters.

l - Length, measured in centimeters.

If we know that w = 6.4\,cm and l = 8.3\,cm, then the perimeter of the rectangle is:

p = 2\cdot (6.4\,cm+8.3\,cm)

p = 29.4\,cm

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter (\Delta p), measured in centimeters, is estimated by differences. That is:

\Delta p = 2\cdot (\Delta w + \Delta l)  (2)

Where:

\Delta w - Uncertainty in width, measured in centimeters.

\Delta l - Uncertainty in length, measured in centimeters.

If we know that \Delta w = 0.2\,cm and \Delta l = 0.2\,cm, then the uncertainty in perimeter is:

\Delta p = 2\cdot (0.2\,cm+0.2\,cm)

\Delta p = 0.8\,cm

The uncertainty in its perimeter is 0.8 centimeters.

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Explain why the sound produced by every vibrating body cannot be heard by us ?​
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Answer:

The human ear is not sensitive to every frequency of sound, rather, our hearing range is from 20Hz to 20,000Hz. This means sound frequencies outside this range are not audible to the human ear, which is why not every sound produced is heard. This is quite a blessing because if we could hear every vibrating body, then our ears would never stop listening: we won't be able to distinguish 'important sounds' from unimportant ones, rendering our ears effectively useless. Perhaps this is why we have a hearing range in the first place: sounds of our friends and our foes lie in 20Hz-20kHz range.

5 0
3 years ago
Consider the following four objects: a hoop, a solid sphere, a flat disk, a hollow sphere. Each of the objects has mass M and ra
Svetlanka [38]

Answer:

Hoop.

Explanation:

The angular acceleration performed at a given torque:

\alpha = \frac{\tau}{I}

The moments of inertia of each element are described below:

Hoop

I = M\cdot R^{2}

Solid sphere

I = \frac{2}{5}\cdot M \cdot R^{2}

Flat disk

I = \frac{1}{2}\cdot M \cdot R^{2}

Hollow sphere

I = \frac{2}{3}\cdot M \cdot R^{2}

The greater the moment of inertia, the greater the torque to obtain the same angular acceleration. Therefore, the hoop requires the largest torque to receive the same angular acceleration.

8 0
3 years ago
The average mass of a car in the US is 1.440 x 10^6 g. Express this mass in kg.
wolverine [178]

Answer:

Average mass of acar in the US (in kg) = 1440 kg

Explanation:

Average mass of a car in the US (in g) = 1.440 × 10⁶ g

Mass in kg:

\rm 1 \: g =  {10}^{ - 3}  \: kg \\  \\  \rm 1.440 \times 10^6 \ g = 1.440 \times 10^6  \times  {10}^{ - 3} \ kg \\  \\  \rm = 1.440 \times 10^{6 - 3} \ kg \\  \\  \rm = 1.440 \times 10^3 \ kg \\  \\ \rm = 1.440 \times 1000 \ kg \\  \\ \rm = 1440 \ kg

5 0
3 years ago
Suppose we hang a heavy ball with a mass 13 kg (so the weight is ) from a steel wire 3.9 m long that is 3.1 mm in diameter (radi
Katena32 [7]

Answer:

1.635×10^-3m

Explanation:

Young modulus is the ratio of the tensile stress of a material to its tensile strain.

Young modulus = Tensile stress/tensile strain

Tensile stress = Force/Area

Given force = 130N

Area = Πr² = Π×(1.55×10^-3)²

Area = 4.87×10^-6m²

Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²

Tensile strain = extension/original length

Tensile strain = e/3.9

Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²

2×10¹¹N/m² = 8.39×10^7/{e/3.9)}

2×10¹¹ = (8.39×10^7×3.9)/e

2×10¹¹e = 3.27×10^8

e = 3.27×10^8/2×10¹¹

e = 1.635×10^-3m

The stretch of the steel wire will be

1.635×10^-3m

7 0
3 years ago
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