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Stolb23 [73]
3 years ago
14

A tree on a hillside casts a shadow 215 ft down the hill. If the angle of inclination of the hillside is 22???? to the horizonta

l and the angle of elevation of the sun is 52????, find the height of the tree.

Physics
1 answer:
krek1111 [17]3 years ago
3 0

Answer:

H=174.168ft

Explanation:

See diagram below.

First, calculate y using sine function:

sin(22)=\frac{y'}{215} \\y'=215sin(22)=80.540ft

Then, calculate x using cosine function:

cos(22)=\frac{x}{215} \\x=215cos(22)=199.345ft

Now use tangent to calculate y:

tan(52)=\frac{y+y'}{x} \\y=x*tan(52)-y'\\y = 174.168ft.

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6 0
3 years ago
Read 2 more answers
Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3
Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

k_{1}=20\ N/m

k_{2}=30\ N/m

k_{3}=15\ N/m

k_{4}=20\ N/m

k_{5}=35\ N/m

According to figure,

k_{2} and k_{3} is in series

We need to calculate the equivalent

Using formula for series

\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}

k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}

Put the value into the formula

k=\dfrac{30\times15}{30+15}

k=10\ N/m

k and k_{4} is in parallel

We need to calculate the k'

Using formula for parallel

k'=k+k_{4}

Put the value into the formula

k'=10+20

k'=30\ N/m

k_{1},k' and k_{5} is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}

Put the value into the formula

k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}

k_{eq}=8.93\ N/m

Hence, The equivalent stiffness of the string is 8.93 N/m.

3 0
3 years ago
A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16
erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle \phi = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

V =\pi D( \frac{N}{60} )

V = 3.14 (0.034) \frac{(1650)}{60}

V = 2.93 \frac{m}{s}

Form factor for the pinion gear is

Y = 0.303

Now

K_{v} = \frac{6.1 +0.303}{6.1} = 1.049

Force on gear tooth

F = \frac{P}{V}

F = \frac{1200}{2.93}

F = 408.73 N

Now the bending stress is given by the formula

\sigma = \frac{K_{v} F}{m b y}

\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}

\sigma = 35.38 M pa

This is the value of bending stress on the pinion

8 0
3 years ago
A 20-N force acts on an object with a mass of 2.0kg. what is the objects acceleration?
ale4655 [162]

Answer:

10 m/s^2

Explanation:

Equation: F = ma.

a = acceleration

m = mass

F = force

Because we are trying to find acceleration instead of force we want to rearrange the equation to solve for a which is F/m = a.

F = 20

m = 2

a = ?

a = F/m

a = 20/2

a = 10 m/s^2

7 0
3 years ago
Read 2 more answers
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