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3 years ago

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Help please ASAP !!!

1 answer:

mezya [45]3 years ago

7 0

What’s the weight and how high is the clif

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a baseball pitcher throws a fastball at 42 meters per second. if the batter is 18 meters from the pitcher, approximately how muc

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T=D/v = 18/42 = 43 seconds

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3 years ago

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2. We know that the Earth is moving around the sun. When we look at the sky we see

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Answer: The Sun, the Moon, the planets, and the stars all rise in the east and set in the west. And that's because Earth spins -- toward the east. ... Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

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3 years ago

Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago

A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and

frozen [14]

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

$m=\sqrt{\dfrac{hP}{KA}}$

For circular fin

$m=\sqrt{\dfrac{4h}{KD}}$

$m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}$

m = 37.08

$\eta_f=\dfrac{tanhmL}{mL}$

$0.65=\dfrac{tanh37.08L}{37.08L}$

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

$\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}$

$\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}$

ε = 23.52

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3 years ago

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Answer:

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Explanation:

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2 years ago