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nexus9112 [7]
3 years ago
12

A surveyor is trying to find the height of a hill . he/she takes a sight on the top of the hill and find that the angle of eleva

tion is 40°. he/she move a distance of 150 metres on level ground directly away from the hill and take a second sight. from this point the angl.e of elevation is 22°. find the height of the
hill​
Engineering
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

height ≈ 60.60 m

Explanation:

The surveyor is trying to find the height of the hill . He takes a sight on the top of the hill and finds the angle of elevation is 40°. The distance from the hill where he measured the angle of elevation of 40° is not known.

Now he moves 150 m on level ground directly away from the hill and take a second sight from this point and measures the angle of elevation as 22°. This illustration forms a right angle triangle. The opposite side of the triangle is the height of the hill. The adjacent side of the triangle which is 150 m is the distance on level ground directly away from the hill.

Using tangential ratio,

tan 22° = opposite/adjacent

tan 22° = h/150

h = 150 × tan 22°

h = 150 × 0.40402622583

h = 60.6039338753

height ≈ 60.60 m

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3 years ago
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2 years ago
Timken rates its bearings for 3000 hours at 500 rev/min. Determine the catalog rating for a ball bearing running for 10000 hours
svet-max [94.6K]

Answer:

C₁₀ = 6.3 KN

Explanation:

The catalog rating of a bearing can be found by using the following formula:

C₁₀ = F [Ln/L₀n₀]^1/3

where,

C₁₀ = Catalog Rating = ?

F = Design Load = 2.75 KN

L = Design Life = 1800 rev/min

n = No. of Hours Desired = 10000 h

L₀ = Rating Life = 500 rev/min

n₀ = No. of Hours Rated = 3000 h

Therefore,

C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3

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1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by
zvonat [6]

Answer:

299.36 feet

Explanation:

To \ find  \   the  \ distance \  of  \ the  \ ball \  from  \ the \ home  \ plate.  \\ \\ From  \ the  \ given  \ information:

Height \ h = 4 \ ft

Initial \ speed \ V_o = 98 \ ft/s ec

The  \ angle \  \theta = 45^0

Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s

U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}

U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}

So;

S_y = u_y t - \dfrac{1}{2}gt^2

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By solving:

t_1 = 4.32 \ sec

Thus;

horizontal \ distance = U_x t

= \dfrac{98}{\sqrt{2}}\times 4.32

\mathbf{=299.36  \ feet}

\mathbf{Thus \ , the  \  distance \ from \  the  \ home  \ plate \  =  \ 299.36  \ feet}

5 0
2 years ago
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