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Arisa [49]
3 years ago
13

The discharge of a stream is Choose one: likely to decrease downstream in arid regions and increase downstream in temperate regi

ons. typically lower in spring than during summer. constant for the length of the stream. calculated by dividing its cross-sectional area by its velocity.
Physics
1 answer:
Sauron [17]3 years ago
3 0

Answer:

<em>likely to decrease downstream in arid regions and increase downstream in temperate regions</em>

<em></em>

Explanation:

Arid regions are is a region with a severe lack of water, usually to the extent that affect the organisms living in the region. Arid regions are characterized by a very low depth of rainfall per year. Temperate region on the other hand experience more distinct seasonal change and wider temperature change. Temperate regions get a fairly large amount of rainfall per year.

In arid regions, the soil is very dry, and the rate of infiltration and percolation is high relative to the amount of rainfall available. The effect is that more water is infiltrated into the soil as you move downstream, leading to a decrease in the discharge of a stream as you move downstream. Most temperate region have soils that are usually saturated in the peak of the rainfall season, leading to a greater stream discharge as you move downstream.  

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5 0
2 years ago
At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar
swat32

Answer:

C=1,25\cdot 10^{5} kJ/^{\circ}C

Explanation:

First of all let's define the specific molar heat capacity.

C = \frac{-Q}{n\cdot \Delta T} (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system  

Now, we can find n with the molar mass (M) the mass of the compound (m).

n=\frac{m}{M}=6.95\cdot 10^{-3} moles      

Using (1) we have:

C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}

C=1,25\cdot 10^{5} kJ/^{\circ}C

I hope it helps!

6 0
3 years ago
A sound with an intensity greater than_____ dB can cause damage to human ears.
sergiy2304 [10]

85 decibels or higher can cause damage to the human ear. Decibels is the value of the measurement of sound. Hearing loss can be caused by noise coming from a loud sound. Noise induced hearing loss maybe permanent loss or temporary loss. Examples of activities that can give you a noise induced hearing loss is target shooting, listening to music in your earphones that have a high volume, and using lawnmowers. 

5 0
3 years ago
If the mass of the sun is 1x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____
nasty-shy [4]

If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 2, 6, and 75, and all planets will orbit the sun successfully.

If the mass of the sun is 2x, at least one planet will fall into the habitable zone. if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully.

If the mass of the sun is 3x, at least one planet will fall into the habitable zone if I place a planet in orbits 672, and 7 and all planets will orbit the sun successfully.

8 0
2 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
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