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Amiraneli [1.4K]
3 years ago
14

1. The 10-min unithydrograph for a 2.25 km2 urban catchment is given by:(a) estimate the run off hydrograph for a 10-min rainfal

l excess of 5.5 cm, (b) estimate the run off hydrograph for a 30-min rainfall. What is the rainfall excess
Physics
1 answer:
vivado [14]3 years ago
5 0

Answer:

A ) Find the direct runoff depth; d =V A 22,698 _ 2.25x106 =0.01m =1cm The above result shows that the given hydrograph qualifies as a unit hydrograph. .

Explanation:

Brainlyist

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3 years ago
A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of gla
Diano4ka-milaya [45]

Complete Question

A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,  

A :

a force repels the glass out of the capacitor.  

B :

a force attracts the glass into the capacitor.    

C :

no force acts on the glass.      

D :

a net charge appears on the glass.      

E :

the glass makes the plates repel each other.

Answer:

The correct option is B

Explanation:

Generally when the glass dielectric is slowly inserted between the plated,

The positive plate of the capacitor will induce a negative charge on the glass while the negative  plate of the capacitor will induce a positive charge on glass which a electric field that posses an electric force that will attract the glass

3 0
3 years ago
Why does the mass spectrum of Br2 contain three signals whose heights are almost in the ratio of 1:2:1? What are the origins of
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One major disadvantage of breeder reactors is that they can do which of the following? Select one:a.explodeb.leak radioactivityc
zimovet [89]

definition of breeder reactors.

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8 0
2 years ago
A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should th
Pavel [41]

Answer:

The block's mass should be 3m

Explanation:

Given:

Cart with mass m

From the conservation of energy before mass is added,

  \frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}

Where A = amplitude of spring mass system, k = spring constant

  A = v\sqrt{\frac{m}{k} }

Now new mass M is added to the system,

   \frac{1}{2} (m +M ) v^{2}  = \frac{1}{2}  k A^{2}

  A = v \sqrt{\frac{m +M }{k} }

Here, given in question frequency is reduced to half so we can write,

   f' = \frac{f}{2}

Where f = frequency of system before mass is added, f' = frequency of system after mass is added.

        \omega ' = \frac{\omega}{2}

\sqrt{\frac{k}{m +M} }  = \frac{\sqrt{\frac{k}{m} } }{2}

   \frac{k}{m +M } = \frac{k}{4m}

   M = 3m

Therefore, the block's mass should be 3m

8 0
3 years ago
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