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2 years ago

6

4) A steel tape is placed around the earth at the equator when the temperature is 0 C. What will the clearance between the tape

and the ground (assumed to be uniform) be if the temperature of the tape rises to 30 C. Neglect the expansion of the earth (the radius of the earth is 6.37 X 106 m)

1 answer:

snow_lady [41]2 years ago

4 0

Answer:

2102.1 m

Explanation:

Temperature at the equator = 0⁰

Radius of the earth = 6.37x10⁶

Required:

We how to find out what the clearance between tape and ground would be if temperature increases to 30 degrees.

Final temperature = ∆T = 303-273 = 30

S = 11x10^-6

The clearance R = Ro*S*∆T

=6.37x10⁶x 11x10^-6x30

= 2102.1m

Or 2.102 kilometers

Thank you

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Answer:

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1. The construction process begins with which of the following stages?

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Answer:

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3 years ago

An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e

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Answer:

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The automobile weight is 2500 lbf.

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The initial elevation is of 5183 ft.

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

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Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:

$Ep = m*g*(h_2 - h_1)\\ W = m*g$

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.

$h_2$ is the final elevation and $h_1$ is the initial elevation.

Replacing W in the Ep equation

$Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\$

Finally, the next step is to replace the variables of the problem.

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3 years ago

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ivann1987 [24]

Answer:

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1 year ago

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. A

irina [24]

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.

So; we can say :

$D_{pursuit} =D_{police}$

By using the second equation of motion to find the distance S;

$S= ut + \dfrac{1}{2}at^2$

$D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)$

$D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)$

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$D_{police} = 0.98*(t^2 + 144 + 24t)$

$D_{police} = 0.98t^2 + 141.12 + 23.52t$

Recall that:

$D_{pursuit} =D_{police}$

$(187.8+15.65 \ t-1.525 t^2)= 0.98t^2 + 141.12 + 23.52t$

$(187.8 - 141.12) + (15.65 \ t - 23.52t) -( 1.525 t^2 - 0.98t^2) = 0$

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t = -6.16 OR t = 3.02

Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs

From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;

Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s

Total time required for the police car to over take the automobile = 15.02 sec

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