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HACTEHA [7]
2 years ago
5

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

Physics
1 answer:
Elanso [62]2 years ago
6 0

The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

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An automobile is traveling away from Jill and towards Jack. The horn is honking, producing a sound wave.
wlad13 [49]

Answer:

a. The sound will travel at the speed to both Jill and Jack

b. Jack

Explanation:

a. Doppler effect describes how the frequency of a sound wave changes with regards to an observer that has relative motion to the sound source;

The Doppler effect is given by the following formula;

For a sound that is moving away, as observed by Jill, we have;

f' = \dfrac{(v - v_0)}{(v + v_s)} \cdot f =  \dfrac{v }{(v + v_s)} \cdot f

For approaching sound, as observed by Jack, we have;

f' = \dfrac{(v + v_0)}{(v - v_s)} \cdot f =  \dfrac{v }{(v - v_s)} \cdot f

Where;

f = The sound wave's actual frequency

f' = The frequency of the moving sound to the observer

v = The speed of the sound wave

v₀ = The observer's velocity = 0

v_s = The velocity of the source (the automobile honking) of the sound wave

From the above equation, we have that the speed of sound, 'v', is the same to both the source, and the observer although the frequency, and therefore, the wavelength of the sound alternatively increases or decreases

b. From the Doppler effect equation, the person who will hear the highest frequency is given by the formula for the frequency when the sound is approaching the observer, which has the lower denominator

Therefore, given that the automobile is travelling towards Jack, jack will hear the higher frequency

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3 years ago
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On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

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mixer [17]

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Where x is displacement from equilibrium

K = spring constant

Therefore X(t) = Acos(2pit/T)

X(t)= Acos(wt)

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