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tino4ka555 [31]
3 years ago
11

A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through a

n angle θ . How should θ be chosen so that the gutter will carry the maximum amount of water?
Physics
1 answer:
GarryVolchara [31]3 years ago
7 0

Answer:

60

Explanation:

The volume of the water carried will be proportional to the cross-sectional area. The cross-section is a trapezoid with height 10sint, where t represents theta. The top of the trapezoid is 10+(2)(10cost), i.e. 10 + 20cost. The base of the trapezoid is 10.

Area of trapezoid = (average of bases) times (height)

= ([10 + (10 +20cost)] / 2 ) 10 sint

= (10 + 10cost)(10sint)

= 100sint + 100sintcost, call this A(t). You need to maximize. So differentiate and set equal to zero.

dA/dt = -100sin(t)^2+100cos(t)+100cos(t)^2 = 0, divide by 100:

-sin(t)^2 + cos(t) + cos(t)^2 = 0, replace sin^2 by 1-cos^2

2cos(t)^2 + cos(t) - 1 = 0, factor

(1+cos(t))(2cos(t)-1)=0, so  

cos(t)=1, t=0 that give a min (zero area) not a max, or

cos(t) = 1/2, so t=60 degrees. This gives the max.

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Is the visible light color spectrum a small segment of the electromagnetic spectrum?
egoroff_w [7]

Answer:

Yes

Explanation:

The visible light color spectrum only makes up a small segment of the EM spectrum.

7 0
3 years ago
A soccer ball of mass 0.4 kg is moving horizontally with a speed of 20 m/s when it is kicked by a player. The kicking force is s
amid [387]

Answer:

v_{f} = 74 m/s, F = 230 N

Explanation:

We can work on this exercise using the relationship between momentum and moment

        I = ∫ F dt = Δp

bold indicates vectors

we can write this equations in its components

X axis

       Fₓ t = m ( -v_{xo})

Y axis  

        t = m (v_{yf} - v_{yo})

in this case with the ball it travels horizontally v_{yo} = 0

Let's use trigonometry to write the final velocities and the force

        sin 30 = v_{yf} / vf

        cos 30 = v_{xf} / vf

        v_{yf} = vf sin 30

        v_{xf} = vf cos 30

         sin40 = F_{y} / F

         F_{y} = F sin 40

         cos 40 = Fₓ / F

         Fₓ = F cos 40

let's substitute

      F cos 40 t = m ( cos 30 - vₓ₀)

      F sin 40 t = m (v_{f} sin 30-0)

we have two equations and two unknowns, so the system can be solved

        F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)

        F sin 40 0.1 = 0.4 v_{f} sin 30

we clear fen the second equation and subtitles in the first

         F = 4 sin30 /sin40     v_{f}

         F = 3.111 v_{f}

        (3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80

        v_{f} (3,111 cos 40 -4 cos30) = - 80

        v_{f} (- 1.0812) = - 80

        v_{f} = 73.99

        v_{f} = 74 m/s

now we can calculate the force

          F = 3.111 73.99

          F = 230 N

6 0
3 years ago
6. Applying Explain how scientists see what early galaxies looked like..
svlad2 [7]

Answer:  The younger elliptical and lenticular galaxies had results similar to spiral galaxies like the Milky Way. The researchers found that the older galaxies have a larger fraction of low-mass stars than their younger counterparts.

Explanation:

6 0
2 years ago
A water balloon is hovering directly above the line join points ANB which are 4.6 km apart if the angles of elevation to the bal
Anna35 [415]

Answer:

Drawing the triangle:

H / x = tan 52.2 = 1.29

H / (4.6 - x) = tan 28.8 = .550

H = 1.29 x

H = .55 * 4.6 - .55 x

1.84 x = 2.53        combining equations

x = 1.38

4.6 - 1.38 = 3.22

Total base of triangle = 1.38 + 3.22 = 4.6

H / x = tan 52,2 = 1.29

H = 1.29 * 1.38 = 1.78 height of triangle

Check:

1.78 / 3.22 = tan 28.9    

This agrees with the given value of 28.8

7 0
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