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3 years ago

15

A hammer can be used to see how a mineral breaks. If you observe square chunks of the mineral when broken, what can you conclude

? (1 point)

1 answer:

jeyben [28]3 years ago

6 0

Answer:

it can not break that easily

Explanation:

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Given frequency, what is the formula for the period of a wave?

Veronika [31]

Answer:

f = c / λ = wave speed c (m/s) / wavelength λ (m). The formula for time is: T (period) = 1 / f (frequency). λ = c / f = wave speed c (m/s) / frequency f (Hz). The unit hertz (Hz) was once called cps = cycles per second.

Explanation:

7 0

3 years ago

Natalie solved the problem below incorrectly. Identify his error and justify your reasoning. −19+3x−11+2x=2 5x−19−11=2 5x−30=2 5

arsen [322]

5x-30=2
5x=2+30 (not -28) when the -30 is brought over to the RHS, 30 should be added to 2 instead of subtracted

hence, 5x=32
x = 6.4

5 0

3 years ago

For the speed equation along centerline of a diffuser, calculate the fluid acceleration along the diffuser centerline as a funct

Marrrta [24]

Answer:

$a = v\cdot \frac{dv}{dx}$ , $v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}$ , $\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}$

Explanation:

Let suppose that fluid is incompressible and diffuser works at steady state. A diffuser reduces velocity at the expense of pressure, which can be modelled by using the Principle of Mass Conservation:

$\dot m_{in} - \dot m_{out} = 0$

$\dot m_{in} = \dot m_{out}$

$\dot V_{in} = \dot V_{out}$

$v_{in} \cdot A_{in} = v_{out}\cdot A_{out}$

The following relation are found:

$\frac{v_{out}}{v_{in}} = \frac{A_{in}}{A_{out}}$

The new relationship is determined by means of linear interpolation:

$A (x) = A_{in} +\frac{A_{out}-A_{in}}{L}\cdot x$

$\frac{A(x)}{A_{in}} = 1 + \left(\frac{1}{L}\right)\cdot \left( \frac{A_{out}}{A_{in}}-1\right)\cdot x$

After some algebraic manipulation, the following for the velocity as a function of position is obtained hereafter:

$\frac{v_{in}}{v(x)} = 1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1\right) \cdot x$

$v(x) = \frac{v_{in}}{1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x}$

$v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}$

The acceleration can be calculated by using the following derivative:

$a = v\cdot \frac{dv}{dx}$

The derivative of the velocity in terms of position is:

$\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}$

The expression for acceleration is derived by replacing each variable and simplifying the resultant formula.

8 0

3 years ago

Read 2 more answers

Which of the following animals can be found in the locality

erastovalidia [21]

Answer:

c frog

Explanation:

frog have chance that can lived in the locality or in public place like river,lake,farm and other place that they can be lived and hunt foods. In the picture bear, elephant and tiger are can't lived in the locality that have many population of people. little percent of changes that they can hunts foods or lived in the public/locatlity place.

7 0

3 years ago

Copper spheres of 20-mm diameter are quenched by being dropped into a tank of water that is maintained at 280 K . The spheres ma

Ivenika [448]

Answer:

The height of the water is 1.25 m

Explanation:

copper properties are:

Kc=385 W/mK

D=20x10^-3 m

gc=8960 kg/m^3

Cp=385 J/kg*K

R=10x10^-3 m

Water properties at 280 K

pw=1000 kg/m^3

Kw=0.582

v=0.1247x10^-6 m^2/s

The drag force is:

$F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}$

The bouyancy force is:

$F_{B} =V*p_{w} *g$

The weight is:

$W=V*p_{c} *g$

Laminar flow:

$v_{T} =\frac{p_{c}-p_{w}*g*D^{2} }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2} }{18*0.00143} =1213.48 m/s$

Reynold number:

$Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1$

Not flow region

For Newton flow region:

$v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w} }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} } =2.186m/s$

$Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4$

$Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31$

$Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3} )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99$

$Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K$

$\frac{T-T_{c} }{T_{w}-T_{c} } =e^{-t/T} \\T=\frac{m_{c}C_{p} }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s$

$e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m$

8 0

3 years ago