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2 years ago

Select the appropriate shape for the given volume form

1 answer:

3 0

Fist one is a cylinder
the second, i believe is a sphere
the third is a rectangular prism
and the last is the same as the first, a cylinder

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Problem 12: A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong w

avanturin [10]

Answer:

The deformation in the pole due to force is 0.70 mm.

Explanation:

Given that,

Height = 20.0 m

Diameter = 4.00 cm

Force = 1100 N

We need to calculate the area

Using formula of area

$A=\pi\times r^2$

$A=\pi\times(2.00\times10^{-2})^2$

$A=0.00125\ m^2$

$A=1.25\times10^{-3}\ m^2$

We need to calculate the deformation

Using formula of deformation

$\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)$

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

$\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)$

$\Delta x=0.000704\ m$

$\Delta x=7.04\times10^{-4}\ m$

$\DElta x=0.70\ mm$

Hence, The deformation in the pole due to force is 0.70 mm.

3 0

4 years ago

A light, inextensible cord passes over a light,

Musya8 [376]

Answer: T = 93 N

Explanation:

Assuming the pulley is ideal meaning frictionless as mentioned and also negligible mass.

ΣF = Σma

Mg - mg = Ma + ma

a = g(M - m) / (M + m)

Now looking only at the larger mass as it falls

Mg - T = Ma

T = Mg - Ma

T = Mg - Mg(M - m) / (M + m)

T = Mg(1 -(M - m) / (M + m))

T = 16(9.8)(1 - (16 - 6.7) / (16 + 6.7))

T = 93 N

or looking only at the smaller mass

T - mg = ma

T = m(g + a)

T = m(g + g(M - m) / (M + m))

T = mg(1 + (M - m) / (M + m))

T = 6.7(9.8)(1 + (16 - 6.7) / (16 + 6.7))

T = 93 N

3 0

3 years ago

I need help with this

polet [3.4K]

Exothermic because the heat make part of the result

7 0

3 years ago

Determine the centroid of the shaded area shown in figure 2. Determine the moment of inertia about y-axis of the shaded area sho

Nady [450]

Answer:

centroid: (x, y) = (81.25 mm, 137.5 mm)
I = 8719.31 mm^2 for unit mass

Explanation:

Finding the desired measures requires we know a differential of area. That, in turn, requires we have a way to describe a differential of area. Here, we choose to use a vertical slice, which requires we know the area boundaries as a function of x.

The upper boundary is a line with a slope of 125/156.25 = 0.8, and a y-intercept of 125. That is, ...

y1 = 0.8x +125

The lower boundary is given in terms of y, but we can solve for y to find ...

100x = y^2

y2 = 10√x

Then our differential of area is ...

dA = (y1 -y2)dx

The centroid is found by computing the first moment about the x- and y-axes, and dividing those values by the area of the figure.

The area will be ...

$\displaystyle A=\int_0^{156.25}{dA}=\int_0^{156.25}{(y_1-y_2)}\,dx$

The y-coordinate of the centroid is ...

$\displaystyle \overline{y}=\dfrac{S_x}{A}=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}}\,dA=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}(y_1-y_2)}\,dx=137.5$

Similarly, the x-coordinate is ...

$\displaystyle \overline{x}=\dfrac{S_y}{A}=\dfrac{1}{A}\int_0^{156.25}{x}\,dA=\dfrac{1}{A}\int_0^{156.25}{x(y_1-y_2)}\,dx=81.25$

That is, centroid coordinates are (x, y) = (81.25, 137.5) mm.

The moment of inertia is the second moment of the area. If we normalize by the "mass" (area), then the integral looks a lot like the one for $\overline{x}$ , but multiplies dA by x^2 instead of x.

The attachment shows that value to be ...

I ≈ 8719.31 mm^2 (normalized by area)

The area is 16276.0416667 mm^2, if you want to "un-normalize" the moment of inertia.

7 0

3 years ago

Calculate the mean free path of air molecules at a pressure of 7.00×10^−13 atm and a temperature of 303 K . (This pressure is re

Grace [21]

Answer:

82.8986 km

Explanation:

Given:

Pressure = 7.00×10⁻¹³ atm

Since , 1 atm = 101325 Pa

So, Pressure = 7.00×10⁻¹³×101325 Pa = 7.09275×10⁻⁸ Pa

Radius = 2.00×10⁻¹⁰ m

Diameter = 4.00×10⁻¹⁰ m (2× Radius)

Temperature = 303 K

The expression for mean free path is:

$\lambda (Mean\ free\ path)=\frac {K (Boltzmann\ Constant)\times Temperature}{\sqrt {2}\times \pi\times (Diameter)^2\times Pressure}$

Boltzmann Constant = 1.38×10⁻²³ J/K

So,

$\lambda (Mean\ free\ path)=\frac {1.38\times 10^{-23}\times 303}{\sqrt {2}\times \frac {22}{7}\times (4.00\times 10^{-10})^2\times 7.09275\times 10^{-8}}$

4 0

4 years ago