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solniwko [45]
3 years ago
15

A river with large, smooth rocks on the bottom, small and medium sized sediment is moving downstream. Which conditions resulted

in the erosion of larger rocks, like the one circled in the image? Check all that apply. low gradient, low velocity low gradient, high velocity low gradient, high volume high gradient, low velocity high gradient, high velocity
Mathematics
2 answers:
Tamiku [17]3 years ago
4 0

Answer:

low gradient, high volume

high gradient, low velocity

high gradient, high velocity

Step-by-step explanation:

zimovet [89]3 years ago
3 0

Answer:

low gradient, high volume

high gradient, low velocity

high gradient, high velocity

Step-by-step explanation:

On E2020

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Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

7 0
2 years ago
Find the volume of the cube 1/3 km 1/3 km 1/3 km
guapka [62]

Answer:

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Step-by-step explanation:

As The volume of cube =the cube(^3) of the sides.

7 0
3 years ago
At the beginning of year 1, Josie invests $400 at an annual compound interest
zvonat [6]

Answer:

The explicit formula that can be used is A=\$400(1.05)^{2}

The account's balance at the beginning of year 3 is A=\$441  

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

t=2\ years\\ P=\$400\\ r=0.05\\n=1  

substitute in the formula above  

A=\$400(1+\frac{0.05}{1})^{1*2}  

A=\$400(1.05)^{2}

A=\$441  

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3 years ago
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