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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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A projectile is fired at an upward angle of 35.6° from the top of a 208-m-high cliff with a speed of 170-m/s. What will be its s
sdas [7]

Answer:

Final velocity is 181.61 m/s at angle 40.44° below horizontal.

Explanation:

Initial horizontal velocity = 170 cos 35.6 = 138.23 m/s

Final horizontal velocity = 138.23 m/s

Considering vertical motion of projectile:

Initial vertical velocity, u = 170 sin 35.6 = 98.96 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = -208 m

We have v² = u² + 2as

Substituting

             v² = 98.96² + 2 x -9.81 x -208

             v = 117.79 m/s

Final velocity,

            v=\sqrt{138.23^2+117.79^2}=181.61m/s

            \theta =tan^{-1}\left ( \frac{117.79}{138.23}\right )=40.44^0

Final velocity is 181.61 m/s at angle 40.44° below horizontal.

5 0
4 years ago
What is absolute zero? What is the temperature of absolute zero on the Kelvin and Celsius scales?
salantis [7]

Answer:

Absolute zero = 0 K or - 273°C

Explanation:

Absolute zero :

 When the entropy and enthalpy of the ideal system reach at the minimum value then the temperature at that condition is known as absolute zero condition.

Absolute temperature is the minimum temperature in the temperature scale.The value of absolute zero is 0 K.

We know that

\dfrac{C-0}{100}=\dfrac{K-273}{100}=\dfrac{F-32}{180}

F=Temperature in Fahrenheit scale

K=Temperature in Kelvin scale

C=Temperature in degree Celsius scale

When  K = 0

\dfrac{C-0}{100}=\dfrac{K-273}{100}

\dfrac{C-0}{100}=\dfrac{0-273}{100}

C= - 273°C

Absolute zero = 0 K or - 273°C

6 0
3 years ago
Read 2 more answers
Provide a brief summary of your selected animal, and describe why it is not extinct.
mina [271]
I'm going to say a Human.



In one simple statement, Humans have survived all of earth's devastation because of evolution. At one point it is believed that the earth had changing conditions including that some of the food that we hunted in waters became extinct. Then, as lack of food grew, We gained the ability to walk on land. As this happened, we began to adapt to our surroundings, including gaining the ability to walk on two legs and use tools
4 0
3 years ago
Read 2 more answers
In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6
Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

         \lambda = \dfrac{h}{mv}

         1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}

         v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}

              v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

     E = \dfrac{1}{2}mv^2

     E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2

            E = 2.5 x 10⁻¹⁴ J

8 0
3 years ago
Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at
dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

   The amplitude of the lateral  force is  F = 25 \  N

   The frequency is   f = 1 \  Hz

   The mass of the bridge per unit length is  \mu  =  2000 \  kg /m

    The length of the central span is  d =  144 m

     The oscillation amplitude of the section  considered at the time considered is  A = 75 \ mm =  0.075 \  m

      The time taken for the undriven oscillation to decay to \frac{1}{e}  of its original value is  t = 6T

Generally the mass of the section considered is mathematically represented as

            m =  \mu  *  d

=>        m =  2000 * 144

=>        m =  288000 \ kg

Generally the oscillation amplitude of the section after a  time period  t is mathematically represented as

                 A(t) = A_o e^{-\frac{bt}{2m} }

Here b is the damping constant and the A_o is the amplitude of the section when it was undriven

So from the question  

               \frac{A_o}{e}  = A_o e^{-\frac{b6T}{2m} }

=>            \frac{1}{e}  =e^{-\frac{b6T}{2m} }

=>          e^{-1} =e^{-\frac{b6T}{2m} }

=>           -\frac{3T b}{m}  =  -1

=>         b  = \frac{m}{3T}

Generally the amplitude of the section considered is mathematically represented as

           A =  \frac{n * F }{ b *  2 \pi }

=>       A =  \frac{n * F }{ \frac{m}{3T}  *  2 \pi }

=>       n =  A  *  \frac{m}{3}  *  \frac{2\pi}{25}

=>       n = 0.075 *  \frac{288000}{3}  *  \frac{2* 3.142 }{25}

=>       n = 1810 \ people

3 0
3 years ago
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