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Sedaia [141]
2 years ago
5

A shop-vac is capable of pulling in air at a rate of 210 ft^3/min. what is the rate of the vacuum's air flow in l/s?

Physics
1 answer:
Nadya [2.5K]2 years ago
3 0

The correct answer is 99.06 .

Since we know that,

1 ft = 12 inches, (1ft)3 = (12)3inches3

& 1 inch = 2.54 cm ,  (1 inch)3 = (2.54)3cm3

& 1 cm3 = 1ml,

&(1 ml)/1000 = 1L, so (1cm3)/1000 = 1L.

& 1 min. = 60 sec.

Now, 1ft3 = (12)3x(2.54)3/(1000) L

and 1 min = 60 sec

So, 1ft3/min = (12)^3x(2.54)^3/(1000)x(60) L/s

Hence 210 ft3/min = 210x(12^)3x(2.54)^3/(1000)x(60).

= 99.06.

Learn more about rate of the vacuum's air flow here :-

brainly.com/question/24641042

#SPJ4

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Answers can be seen below

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First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

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1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

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k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

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h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

7)

b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

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a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2}  }{M}

9)

v_{o}=\frac{x-\frac{1}{2}*a*t^{2}  }{t}  ; F=\frac{W+uNd}{d*cos(o)}

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h=\frac{E-\frac{1}{2}*m*v^{2}  }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2}  }{\frac{1}{2}m }

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N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

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y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

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