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lilavasa [31]
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)
Engineering
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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A square isothermal chip is of width w 5 mm on a side and is mounted in a substrate such that its side and back surfaces are wel
Lynna [10]

Answer:

a) 0.35 W

b) 5.25 W

Explanation:

Given that

Width of the chip, W = 5 mm

The environment temperature, T(∞) = 15° C

Surface temperature, T(s) = 85° C

The initial convection heat coefficient, h1 = 200 W/m².K

The final convection heat coefficient, h2 = 3000 W/m².K

See attachment for solution

6 0
2 years ago
1. The construction process begins with which of the following stages?
Firdavs [7]

Answer:

c) site preparation

Explanation:

A construction process can be defined as a series of important physical events (processes) that must be accomplished during the execution of a construction project.

Generally, in the construction of any physical asset such as offices, hospitals, schools, stadiums etc, the first step of the construction process is site preparation. Site preparation refers to processes such as clearing, blasting, levelling, landfilling, surveying, cutting, excavating and demolition of all unwanted objects on a piece of land, so as to make it ready for use.

This ultimately implies that, site preparation should be the first task to be accomplished in the construction process.

Hence, the construction process typically begins with site preparation before other activities such as the laying of foundation can be done.

Additionally, construction costs can be defined as the overall costs associated with the development of a built asset, project or property. The construction costs is classified into two (2) main categories and these are; capital and operational costs.

7 0
2 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
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

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.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

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.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
What is the diffrence between a small block and a big block lets see if yall know
ivann1987 [24]

Answer:

The difference in weight and size?

Explanation:

It explains itself :P

5 0
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)

D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)

D_{police} = ut _P + \dfrac{1}{2}at_p^2

where ;

u  = 0

D_{police} =  \dfrac{1}{2}at_p^2

D_{police} =  \dfrac{1}{2}*(1.96)*(t+12)^2

D_{police} = 0.98*(t+12)^2

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

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

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

8 0
2 years ago
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