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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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A ball is thrown with an initial upward velocity of 5 m/s.
miv72 [106K]

Answer:

A ball is thrown at an initial height of 5 feet with an initial upward velocity at 29 ft/s. lets assume that  balls height h (in feet) after t seconds is give by:

<u>h= 5 + 29t -16t^2</u>

Explanation:

h= 5 + 29t -16t^2

 

a time when the ball's height will be 17 ft

 

17 = 5 + 29t -16t2

 

0 = -17 + 5 + 29t -16t2

 

0 = -12 + 29t - 16t2

 

Using the quadratic equation:

 

t = (-29±√(292-(4*(-16)*(-12))))÷2(-16)

 

 = (-29±√(841 - 768))÷(-32)

 

 = (-29±√(73))÷(-32)

 

 = (-29 + 8.544)÷(-32)   or   (-29 - 8.544)÷(-32)

 

 = (-20.456)÷(-32)         or   -37.544÷(-32)

 

 =  0.64                         or   1.17

 

So, the ball is at a height of 17 ft twice: once on the way up after 0.64 seconds and once on the way back down after 1.17 seconds.

6 0
3 years ago
Which is an example of a wedge?<br> a. bowl<br> b. spoon<br> c. fork<br> d. knife
elixir [45]
The answer is d.... Knife.

Hope this helped :)
3 0
3 years ago
A concert loudspeaker suspended high off the ground emits 28.0 W of sound power. A small microphone with a 0.700 cm2 area is 55.
grandymaker [24]

Explanation:

It is known that wave intensity is the power to area ratio.

Mathematically,    I = \frac{P}{A}

As it is given that power is 28.0 W and area is 7 \times 10^{-5} m^{2}.

Therefore, sound intensity will be calculated as follows.

             I = \frac{P}{A}

               = \frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}

                = 0.318 \times 10^{5} W/m^{2}

or,             = 3.18 \times 10^{4} W/m^{2}

Thus, we can conclude that sound intensity at the position of the microphone is 3.18 \times 10^{4} W/m^{2}.

7 0
3 years ago
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.07 Hz. On this tray is an empty cu
Anettt [7]

Answer:

Explanation:

Given

Frequency of SHM is f=2.07\ Hz

Amplitude of SHM is A=3.13\ cm

Cup begins to slip when it overcomes the friction force

Friction force F_s=\mu mg

Applied force F=ma

ma=\mu mg

a=\mu g

and maximum acceleration during SHM is

a=A\omega ^2

a=A(2\pi f)^2

a=3.13\times 10^{-2}\times (2\pi 2.07)^2

a=5.296\ m/s^2

\mu =\frac{a}{g}

\mu =\frac{5.296}{9.8}=0.54

6 0
3 years ago
Read 2 more answers
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