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Zigmanuir [339]
3 years ago
5

An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =

400 K. When placed in atmospheric air at a temperature of T[infinity] = 300 K and moving with a velocity of V = 25 m/s, the average heat flux from the surface to the air is 10,000 W/m2 . If a second object of the same shape, but with a characteristic length of L = 2.5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at T[infinity] = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 5 m/s?.
Physics
1 answer:
goblinko [34]3 years ago
7 0

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of  reynold number for both objects

R_{1}=R_{2}

\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}

\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}

Here, k_{1}=k_{2}

h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}

h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}

Put the value into the formula

h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}

h_{2}=20\ W/Km^2

Hence, The value of the average convection coefficient is 20 W/Km².

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Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of
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Answer:

<em>The frequency changes by a factor of  0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

where f is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

f_{n} = \frac{1}{2\pi } \sqrt{\frac{k}{14m} }

simplifying, we have

f_{n} = \frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }

f_{n} = \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }

if we divide this final frequency by the original frequency, we'll have

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  ÷  \frac{1}{2\pi } \sqrt{\frac{k}{m} }

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  x  2\pi \sqrt{\frac{m}{k} }

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Refer to the figure shown below, which is based on the given figure.

d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.

Part A
From the geometry, obtain
d = X cos(α)                     (1a)
h = X sin(α)                      (1b)

The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α)               (2a)
u = v₀ cos(θ - α)             (2b)

If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0             (3a)
ut = d                                (3b)

Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
0.5(9.8)( \frac{d}{u})^{2} -v_{0} sin(\theta -  \alpha ) \frac{d}{u} - h = 0
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X= \frac{b}{a}

Answer:
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Part B
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Answer:  X = 101.5 m

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