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Temka [501]
3 years ago
14

When the temperature is cold, Tim's tires look under-inflated. This is because the air molecules in the tire ________ causing th

eir kinetic energy to _________ and volume to _______.
The answer is...

A) Slow down, decrease, contract

Because...
When the temperature is cold, Tim's tires look under-inflated. This is because the air molecules in the tire slow down causing their kinetic energy to decrease and volume to contract. slow down, decrease, contract
Physics
1 answer:
lys-0071 [83]3 years ago
3 0
True. The answers to the blanks are slow down, decrease, and contract (in the given order). Molecules move faster when warmer so, because it was cold the molecules in the air slowed down which lessened the kinetic energy and the overall volume of the tires to seem less nd more contracted.
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denis23 [38]

Answer:

False

Explanation:

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5 0
2 years ago
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The resolution of a camera or other optical system is determined by the relationship between what two scales?
devlian [24]

Answer:

d.The wavelength of light and the size of the aperture

Explanation:

<em>The resolution power of an optical system is the smallest distance between two points that the device can distinguish clearly.</em>

It has the following relationship:

r=\frac{\lambda}{2n}

where:

r = minimum resolvable distance

n = numerical aperture

\lambda= wavelength of the light used for viewing

From above mathematical equation it is clear that:

  • Smaller the wavelength better the resolving power
  • Larger the aperture better the resolution

(Note, that smaller the value of "r" the more finer details of the image visible through the device.)

4 0
3 years ago
The circumference of a sphere was measured to be
professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

\phi = 76cm

Error (dr) = 0.5cm

The radius then would be

\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm

And

\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr

PART A ) For the Surface Area we have that,

A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)

Maximum error:

dA = 4(\frac{38}{\pi})(0.5)

dA = \frac{76}{\pi}cm^2

The relative error is that between the value of the Area and the maximum error, therefore:

\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}

\frac{dA}{A} = 0.01315 = 1.31\%

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

V = \frac{4}{3} \pi r^3

V = \frac{4}{3} \pi (\frac{38}{\pi})^3

V = \frac{219488}{3\pi^2}

Therefore the Maximum Error would be,

\frac{dV}{dr} = \frac{4}{3} 3\pi r^2

dV = 2r^2 (2\pi dr)

dV = 4r^2 (\pi dr)

Replacing the value for the radius

dV = 4(\frac{38}{\pi})^2(0.5)

dV = \frac{2888}{\pi^2} cm^3

And the relative Error

\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }

\frac{dV}{V} = 0.03947

\frac{dV}{V} = 3.947\%

3 0
3 years ago
Fluid pressure changes with depth are assumed to be linear. Which statement best explains why this does not hold true for atmosp
ankoles [38]

Answer:

Explanation:

Pressure due to fluid is directly proportional to the depth of fluid, density of the fluid and the value of acceleration due to gravity.

P = h d g

Where, h is the depth, d be the density and g be the acceleration due to gravity.

If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.

4 0
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A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di
Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour          

                                         = Total distance traveled/ total time taken                  

                                         = \frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} }  = \frac{106x}{x+53}

                              26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour

5 0
3 years ago
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