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Elina [12.6K]
2 years ago
13

A 12 cm radius air duct is used to replenish the air of a room 9.2 m x 5.5 m x 4.5 m every 10 min. how fast does the air flow in

the duct?
Physics
1 answer:
mote1985 [20]2 years ago
6 0
In this problem, given the dimensionns of the room and the dimension of the pipe in which air passes through, we can determine the flow of air in the duct. The area of the duct is A = pi r2 where r is equal to 12 cm. we divide this area to time which is 10 minutes, then we get the answer.
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A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this
krok68 [10]

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

8 0
2 years ago
An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the
poizon [28]

Answer:

The value is A  =  0.014 \  m

Explanation:

From the question we are told that

    The mass of the object is  m  =  2.0 \  kg

    The unstressed length of the string is  l  =  0.08 \  m

    The length of the spring when it is  at equilibrium is  l_e = 5.9 \  cm  =  0.059 \  m

      The initial speed (maximum speed)of the spring when given a downward blow v  =  0.30 \  m/s

Generally the maximum speed  of the spring  is mathematically represented as

           u =  A *  w

Here A is maximum height above the floor (i.e the maximum amplitude)

            and w is the angular frequency which is mathematically represented as

       w = \sqrt{\frac{k}{m} }

So

        u =  A *   \sqrt{\frac{k}{m} }

=>      A  =  u *   \sqrt{\frac{m}{k} }

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

           b  =  l -l_e

=>         b  = 0.08 - 0.05 9

=>         b  = 0.021 \  m

Generally at equilibrium position the net force acting on the spring is  

            k *  b  -  mg  =  0

=>         k *  0.021   -   2 * 9.8  =  0

=>        k =  933 \  N/m

So

            A  =  0.30  *   \sqrt{\frac{2}{933} }

=>          A  =  0.014 \  m

8 0
2 years ago
The specification limits for a product are 8 cm and 10 cm. A process that produces the product has a mean of 9.5 cm and a standa
My name is Ann [436]

Answer:

The value of Cpk is 0.83.

Explanation:

Given that,

Upper specification limits = 10 cm

lower specification limits = 8 cm

Mean = 9.5

Standard deviation = 0.2 cm

We need to calculate the process capability

Using formula of Cpk

Cpk=min(\dfrac{USL-mean}{3\times SD}, \dfrac{mean-LSL}{3\times SD})

Put the value into the formula

Cpk=min(\dfrac{10-9.5}{3\times0.2}, \dfrac{9.5-8}{3\times0.2})

Cpk=min(0.83,2.5)

Cpk=0.83

Hence, The value of Cpk is 0.83.

4 0
2 years ago
In a chemical equation, the chemicals that react are considered . In a chemical equation, the chemicals that are produced are co
vovangra [49]

Answer

Hi,

In a chemical equation, chemicals that react are the reactants, while chemicals that are produced are the products/by products. Both sides of the equation must be balanced.

Explanation

When writing a chemical equation, reactants reacts to produce products. For example in the equation for formation of water, hydrogen combines with oxygen as 2H₂ +O₂→2H₂O where the first part before the arrow represent the reactants and the next part after the arrow are the products. Reactants are on the left where as products are on the right.Coefficient 2, in this cases is used for balancing the equation.

Good luck!

4 0
3 years ago
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Use examples to explain how the geosphere interacts with two other of Earth's spheres. Explain the interaction for each using co
Orlov [11]

The geosphere interacts with the hydrosphere when water causes rock to erode. The atmosphere provides the geosphere with heat and energy for erosion, and the geosphere reflects the sun's energy back into the atmosphere.

7 0
3 years ago
Read 2 more answers
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