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Anastasy [175]
4 years ago
5

Assuming that the bridge segments are free to pivot at each intersection point, what is the tension T in the horizontal segment

directly above the point where the object is attached? If you find that the horizontal segment directly above the point where the object is attached is being stretched, indicate this with a positive value for T. If the segment is being compressed, indicate this with a negative value for T. The answer should be in terms of M (mass) and g (gravity).
Physics
1 answer:
Sergio039 [100]4 years ago
4 0
Since the bridge and all segments of it are static, the sum of the torques acting on any portion of the bridge you choose is zero for any pivot <span>point you may choose. See if you can find a rigid portion of the bridge and a wisely chosen pivot to which you can apply this powerful fact.
</span>Consider the triangular portion shown in bold and let x be the pivot. (This choice eliminates the torques due to the tensions in the beams that attach at point x.) Find the torques on this left hand triangle (which can be considered a solid piece because of the connections). Remember that counterclockwise torque is positive. Assume that the horizontal segment above   is being stretched, so that the force that the tension in this segment exerts on the bold triangle is directed to the right. Express the torque in terms of  T, L , and Fp.

Answer in terms of T and L :
Tt = (TL.sqrt 3) / 2
Summation Tx =  -LFp - T sqrt[L^2 - (L/2)^2]
The negative value of the tension shows that the segment is actually under a compressible load.  <span />
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3 years ago
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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

A_{area}=0.140m^{2}\\  p_{air}=1.21 kg/m^{3}\\  m_{mass}=80kg

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

mg=1/2*C*p_{air}*v^{2}*A_{area}\\   v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

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5 0
3 years ago
C Determine the coulomb's force between the electron and the nucleus in
tiny-mole [99]

Answer:

F = 2.3 x 10⁻⁸  N

Explanation:

Given,

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The number of protons present in the nucleus of the hydrogen atom is 1.

The charge of the proton and electron,  q = 1.602 x 10⁻¹⁹ C

Since they have opposite charges, the force between them is attractive,

The coulomb law of force between two charges is given by the formula,

                          <em> F=\frac{1}{4\pi\epsilon_{0}}\frac{qq}{x^{2}}</em>

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Substituting the values in the above equation

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Hence, the force between the nucleus and electron of hydrogen atom is, F = 2.3 x 10⁻⁸  N

6 0
3 years ago
An object moves in a straight path. Its position x as a function of time t is presented by the equation x(t) = at – bt2+c, where
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Answer:

The answer is below

Explanation:

Given that:

x(t) = at – bt2+c

a) x(t) = at – bt2+c

Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:

x(t) = 1.4t - 0.06t² + 50

At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

v=\frac{x(5)-x(0)}{5-0}\\ \\v=\frac{55.5-50}{5-0}=1.1\\ \\v=1.1\ m/s

b) x(t) = 1.4t - 0.06t² + 50

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

v=\frac{x(10)-x(0)}{10-0}\\ \\v=\frac{58-50}{10-0}=0.8\\ \\v=0.8\ m/s

c) x(t) = 1.4t - 0.06t² + 50

At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

The average velocity (v) is given as:

v=\frac{x(15)-x(10)}{15-10}\\ \\v=\frac{57.5-58}{15-10}=0.1\\ \\v=0.1\ m/s

6 0
3 years ago
If the wavelength of a wave of light is 4.57x10^-9 m what is the frequency
alexandr402 [8]

Answer:

6.5646\times 10^{16}~ \text{Hz}

Explanation:

\text{Frequency,}~ f= \dfrac{c}{\lambda} = \dfrac{3\times 10^8 }{4.57 \times 10^{-9}} = 6.5646\times 10^{16}~ \text{Hz}

5 0
3 years ago
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