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2 years ago

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In the diagram, which is demonstrated by arrow 1? A. liquid water evaporating from the surface to the atmosphere B. water vapor

condensing into liquid water in the atmosphere C. liquid water moving from the atmosphere to the surface D. liquid water moving along the surface

1 answer:

Shkiper50 [21]2 years ago

6 0

I think D. liquid water moving along the surfac

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Which events may occur when ocean salinity increases? Check all that apply.

Pavlova-9 [17]

The answer is number two, number four, and number one

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2 years ago

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A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

a)

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

b)

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

$195.948=833sin \phi$

$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

5 0

2 years ago

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

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2 years ago

If a net force applied to an object is equal to zero, then the forces that are acting on an object are considered.

hjlf

Answer:

The mass will accelerate. Balanced Forces: When forces are in balance, acceleration is zero. Velocity is constant and there is no net or unbalanced force. A plane will fly at constant velocity if the acceleration is zero.

Explanation:

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3 years ago

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A kite 100 ft above the ground moves horizontally at a speed of 12 ft/s. at what rate is the angle (in radians) between the stri

Viefleur [7K]

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3 years ago