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postnew [5]
3 years ago
12

A uniform beam of arbitrary, unsymmetrical cross-section and length 2???? is built-in at one end and simply supported in the ver

tical direction at a point halfway along its length. This support, however, allows the beam to deflect freely in the horizontal x direction. For a vertical load W applied at the free end of the beam, calculate bending moments for the sections at A and B.
Physics
1 answer:
leonid [27]3 years ago
5 0

Answer:

 the bending moment will be W from either sides

Explanation:

bending moment= force (load) * perpendicular distance, if I understand the question the distance will be 1/2 of the length

=> f x 1/2(l) =W*1/2(2) =W

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Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

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3 years ago
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3 years ago
Add these measurements, using significant digit rules:
tia_tia [17]

Here we have to add the two measurements given in the question

The measurement values are given as 1.0090 cm and 0.02 cm.we have to  add them on the basis of significant figure rules.

As per the addition rule in terms of significant figures

1-First we have to select the number of significant digits after the decimal point of each quantity.

2-Now we have to remember that during the addition ,the resultant of two quantities will follow the quantity having least number of significant figures after the decimal point.

3-Here we are considering the minimum number of significant figures after the decimal points not the minimum number of significant figures in case of multiplication and division

Now we have to add these two quantities as per the above rule-

         1.0090 cm +0.02 cm

         =1.0290 cm

Here the result  will follow 0.02 which has minimum number of significant figures after the decimal points.

Hence we have to round off the number from 9 of 1.0290

As 9 is  greater than 5 ,so he actual result will be 1.03 cm

       

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2 years ago
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