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Likurg_2 [28]
3 years ago
14

In the diagram, above, marker F is pointing to a __________, which are formed when meanders wear away at a narrow point and a po

ol of water forms.
Physics
2 answers:
castortr0y [4]3 years ago
4 0
The answer is <span>oxbow lake</span>
zubka84 [21]3 years ago
3 0

B. oxbow lake, gl on the program to you smart environmental scientists.

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A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

So, the net pressure acting on the hatch is

p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N

8 0
3 years ago
A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant
atroni [7]

Answer:

The  time constant is  \tau  = 1.265 s

Explanation:

From the question we are told that

     the time take to charge is  t = 2.4 \  s

The mathematically representation for voltage potential of a capacitor at different time is

        V  =  V_o  - e^{-\frac{t}{\tau} }

Where  \tau  is the time constant  

           V_o is the potential of the capacitor when it is full

     So  the capacitor potential will be  100%  when it is full thus  V_o  =100%  =  1  

and from the  question we are told that the  at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

      V  = 0.85

Hence

     0.85 =  1 -  e^{-\frac{2.4}{\tau } }

       - {\frac{2.4}{\tau } }  =  ln0.15

        \tau  = 1.265 s

     

7 0
3 years ago
which of these chemical formulas represents two elements whose atoms have joined together with an ionic bond? A. O2 B. CaO C.NeN
cupoosta [38]
B: CaO
Because Calcium is a Metal and Oxygen is a Non-Metal. Ionic bonds are only formed with a Metal and a Non-Metal element.
3 0
3 years ago
Three parachutists have the following masses: A: 50 kg, B: 40 kg, C: 75 kg Which one has the greatest terminal velocity?
allochka39001 [22]

Answer:

A: 50 kg

Explanation:

5 0
3 years ago
mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi
valina [46]

Answer:

The answer is

"x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs \frac{8}{3} feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\

The source of the hooks law is stable,

16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\

Number \frac{1}{2}  times the immediate speed, i.e .. Damping force

\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2}  \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\

The m^2+m+12=0 and m is an auxiliary equation,

m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \  m2 =\frac{-1 - \sqrt{47i}}{2}

Therefore, additional feature

x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]

Use the form of uncertain coefficients to find a particular solution.  

Assume that solution equation,

x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\

These values are replaced by equation ( 1):

\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\

Going to compare cos3 t and sin 3 t coefficients from both sides,  

The cost3 t is 3A + 3B= 20 coefficients  

The sin 3 t is 3B -3A = 0 coefficient  

The two equations solved:

3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\

Replace the very first equation with the meaning,

3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\

equation is

x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)

The ultimate plan for both the equation is therefore

x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.  

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\

x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t)  +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\

c_2=\frac{-64\sqrt{47}}{141}

x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))

5 0
3 years ago
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