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podryga [215]
3 years ago
6

Can someone help me with a)​

Physics
1 answer:
Gnom [1K]3 years ago
3 0

Answer:

what's your problem. How may I help you Buddy

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If there are 4 molecules of glucose, there will be 24 carbon, 48 hydrogen, and 24 oxygen.
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How do you convert from picoseconds to milliseconds using the conversion factors?
andrew11 [14]

There are 10⁹ picoseconds in 1 Ms

1 picosecond= 10¹² s

1 Ms = 10⁻³ s

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Thus there are 10⁹ picoseconds in 1 Ms

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Que es el ecosistema de los animales
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When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec
igomit [66]

Answer:

The maximum electric power output is P_{max} =1.339*10^{9} \ W

Explanation:

From the question we are told that

        The capacity of the hydroelectric plant is \frac{V}{t}   =  690 \ m^3 /s

         The level at which water is been released is h  =  220 \ m

        The efficiency is  \eta  =0.90

       

The electric power output is mathematically represented as

       P  = \frac{PE_l - PE _o}{t}

Where  PE_l is the potential energy at  level h which is mathematically evaluated as  

          PE_l  =  mgh

and  PE_o  is  the potential energy at ground level which is mathematically evaluated as  

          PE_o  =  mg(0)

         PE_o  =  0

So  

         P  = \frac{mgh}{t}

here  m  =   V *  \rho

where V is volume  and  \rho is density of water whose value is  \rho = 1000 kg/m^3

 So  

         P  = \frac{(\rho * V) * gh}{t}

        P  = \frac{V}{t} * gh \rho

substituting values  

       P  =690 * 9.8 * 220 * 1000

      P  =1.488*10^{9} \ W

The maximum possible electric power output is

           P_{max} = P * \eta

substituting values  

         P_{max} =1.488*10^{9} * 0.90

         P_{max} =1.339*10^{9} \ W

6 0
2 years ago
A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl
Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

Q = C_w BH^{3/2}

H = (\dfrac{Q}{C_wB})^{2/3}

H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}

H = 0.371\ m

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0
3 years ago
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