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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
The velocity of an object is equal to the distance divided by time. The equation is velocity = distance/time. If you wanted to c
Korolek [52]

Answer: C) divide: distance ÷ velocity

Explanation:

The velocity V equation is distance d divided by time t:

V=\frac{d}{t}

If we isolate t we will have:

t=\frac{d}{V}

Hence, the correct option is C: distance divided by velocity.

7 0
3 years ago
Please help me
Grace [21]

Answer:

a. 60 N*s

b. 60 (kg*m)/s

c. 3 m/s

Explanation:

Givens:

m = 20 kg

v_i = 0 m/s

t = 10 s

F = 6 N

a) Impulse:

I = F*t

I = 6 N*10 s

I = 60 N*s

b) Momentum:

p = v*m

F = m(a)

a = F/m

a = 6 N/20 kg

a = 0.3m/s^2

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

p = 3 m/s*20 kg

p = 60 (kg*m)/s

c. Final velocity

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

6 0
3 years ago
A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at
stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft

Also, at any instant t

\Rightarrow l^2=x^2+y^2

differentiate w.r.t.

\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s

5 0
3 years ago
The slope of a good speed Time indicates ?
11Alexandr11 [23.1K]

The answer would be acceleration.

8 0
3 years ago
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