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Maksim231197 [3]
2 years ago
14

A support beam needs to be placed at a 28° angle of elevation so that the top meets a vertical beam 1.6 meters above the horizon

tal floor. the vertical beam meets the floor at a 90° angle. law of sines: startfraction sine (uppercase a) over a endfraction = startfraction sine (uppercase b) over b endfraction = startfraction sine (uppercase c) over c endfraction approximately how far from the vertical beam should the lower end of the support beam be placed along the horizontal floor? 3.0 meters 3.4 meters 3.9 meters 4.4 meters
Physics
1 answer:
k0ka [10]2 years ago
7 0

The distance  from the vertical beam that the lower end of the support beam be placed along the horizontal floor is 3.00 m

<h3>Angle of elevation</h3>

The angle of elevation is the angle to which the eye must be raised in order to see an elevated obect at a height.

Now, we know that a right angle triangle is formed; we can obtain the distance  from the vertical beam that the lower end of the support beam be placed along the horizontal floor from;

tan 28 = 1.6/x

x = 1.6 /tan 28

x = 3.0 meters

Learn more about angle of elevation:brainly.com/question/21137209

#SPJ4

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Answer:

Ben's average speed was twice Debby's average speed.

Explanation:

Ben covered a total distance of 16 miles (10+4+2) and Debby covered 8 miles (3+2+2+1) which is half of what Ben covered. As they both reached the place in the same amount of time it tells us Ben was faster.

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750 kg car zooms away from a red light with an acceleration of 7.8 m/s squared . What is the average net force in Newtons that t
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<h2>5850 N</h2>

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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
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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)

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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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