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tatiyna
3 years ago
10

A uniform steel girder of weight 22KN and of length 12m is lifted off the ground at one end by means of a crane. When the raised

end is 2.0m above the ground, the cable is vertical.
A. Sketch a free-body diagram of the girder in this position
b. Calculate the tension in the cable at this position and the force of the girder on the ground
Physics
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

Explanation:

For free body diagram see attached sheet .

W is weight of steel girder acting at the middle point of its length . T is tension in the cable .

OB = √ ( 12² - 2² )

= 11.83 m .

OC = 11.83 / 2 = 5.915 m

Taking moment of tension T and weight W about point O

W x OC = T x OB

22 x 5.915 = T x 11.83

T = 22 x 5.915 / 11.83

= 11 kN

Considering forces acting in vertical direction and equating forces in opposite direction

T + R = W

R = W - T

= 22 - 11 = 11 KN

So force of grinder on the ground = R

= 11 KN.

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LenKa [72]

The cliff is 2042 ft away.

We know that the speed of sound in air is directly proportional to the absolute temperature.

First convert the Fahrenheit temperature to Celsius;

 °C = 5/9(44.5 - 32)

°C = 6.9 °C

Applying the formula;

V1/V2 = √T1/T2

Where; V1 = velocity of sound in air at  0°C

V2 = Velocity of sound in air at 6.9 °C

1087/V2 = √273/279.9

V2= 1101 ft/s

Given that; V = 2s/t

Where s is the distance of the cliff

t is the time taken

1101 ft/s = 2s/3.71 s

s = 1101 ft/s × 3.71 s/2

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3 years ago
Trey races his bicycle for 192m. A wheel of his bicycle turns 48 times as the bicycle travels this distance. What is the diamete
chubhunter [2.5K]

Answer:

1.27 m

Explanation:

Distance = 192 m

number of rotations = 48

Distance traveled in one rotation = 2 x π x r

Where, r be the radius of wheel.

so, distance traveled in 48 rotations = 48 x 2 x 3.14 x r

It is equal to the distance traveled.

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7 0
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What is the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosp
sineoko [7]

The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

To find the answer, we have to know about the pressure.

<h3>How to find the weight of a column of air?</h3>
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                 P=\frac{F}{A}

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.

  • It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

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  • From this, the value of weight will be,

            F=mg=P*A=1.013*10^5*4.5=4.56*10^5N

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

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brainly.com/question/12830237

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