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Nookie1986 [14]
3 years ago
14

point) A tank in the shape of an inverted right circular cone has height 5 meters and radius 4 meters. It is filled with 4 meter

s of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is δ=1080 kg/m3. Your answer must include the correct units.

Physics
1 answer:
zysi [14]3 years ago
7 0

Answer:

W = 907963.50 J = 907.96 J

Explanation:

Note: Refer to the figure attached

Now, from the figure we have similar triangles ΔAOB and ΔCOD

we have

\frac{5}{4}=\frac{x}{r}

or

r=\frac{4x}{5}

Now, the work done to empty the tank can be given as:

W = \int\limits^4_0 {(5-x)\rho\times g A} \, dx

or

W = \int\limits^4_0 {(5-x)1080\times 9.8 (\pi r^2)} \, dx

or

W = \int\limits^4_0 {(5-x)\times10584\times (\pi (\frac{4x}{5})^2)} \, dx

or

W = 6773.76\pi\int\limits^4_0 {(5-x)x^2)} \, dx

or

W = 6773.76\pi[\frac{5}{3}x^3-\frac{1}{4}x^4]^4_0

or

W = 6773.76\pi[\frac{128}{3}]

or

W = 907963.50 J = 907.96 J

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The string is fixed at two ends with distance 1.5 m. Its mass is 5 g and the tension in the string is 50N and it vibrates on its
liraira [26]

Answer:

a) \lambda=1\ m

b) f=122.47\ Hz

c) \lambda_s=2.8\ m

Explanation:

Given:

distance between the fixed end of strings, l=1.5\ m

mass of string, m=5\ g=0.005\ kg

tension in the string, F_T=50\ N

a)

<u>Since the wave vibrating in the string is in third harmonic:</u>

Therefore wavelength λ of the string:

l=1.5\lambda

\lambda=\frac{1.5}{1.5}

\lambda=1\ m

b)

We know that the velocity of the wave in this case is given by:

v=\sqrt{\frac{F_T}{\mu} }

where:

\mu= linear mass density

v=\sqrt{\frac{50}{(\frac{m}{l}) } }

v=\sqrt{\frac{50}{(\frac{0.005}{1.5}) } }

v=122.47\ m.s^{-1}

<u>Now, frequency:</u>

f=\frac{v}{\lambda}

f=\frac{122.47}{1}

f=122.47\ Hz

c)

When the vibrations produce the sound of the same frequency:

f_s=122.47\ Hz

Velocity of sound in air:

v_s=343\ m.s^{-1}

<u>Wavelength of the sound waves in air:</u>

\lambda_s=\frac{v_s}{f_s}

\lambda_s=2.8\ m

5 0
4 years ago
QUESTION 10
Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

\theta =37.01^{\circ}

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

T_{1}sin(\theta)-T_{2}sin(\theta)=0    (1)

Where:

  • T(1) is the tension due to the rope 1
  • T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

T_{1}cos(\theta)+T_{2}cos(\theta)-W=0   (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)=m_{steel}g

T(1) must be equal to 5479 N, so we have:

cos(\theta)=\frac{m_{steel}g}{2T_{1}}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=0.80

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0
3 years ago
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