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irakobra [83]
3 years ago
12

A tiny but horrible alien is standing at the top of the Eiffel Tower (which is 324324324 meters tall) and threatening to destroy

the city of Paris! A Men In Black agent is standing at ground level, 545454 meters across the Eiffel square, aiming his laser gun at the alien. At what angle, in degrees, should the agent shoot his laser gun? Round your final answer to the nearest tenth. ^\circ ∘ degrees

Mathematics
1 answer:
DedPeter [7]3 years ago
7 0

Answer:

A tiny but horrible alien is standing at the top of the Eiffel Tower (which is 324 meters tall) and threatening to destroy the city of Paris! A Men In Black agent is standing at ground level, 54 meters across the Eiffel square, aiming his laser gun at the alien. At what angle, in degrees, should the agent shoot his laser gun? Round your final answer to the nearest tenth

The agent shoot his laser gun at 80.5 degree.

Step-by-step explanation:

Given:

Height at which the horrible alien is standing = 324 m

Distance from the (tower) where the agent is = 54 m

According to the question:

Men In Black agent is aiming his laser gun to shoot the horrible alien.

Let the angle at which the agent shoot his gun be " Ф "

As from the situated depicted we can draw a right angled triangle.

Using trigonometric ratios:

tan (Ф) = opposite/ adjacent

In triangle OPQ .

The adjacent side = 54 m

The opposite side = 324 m

⇒ Plugging the values.

⇒ tan(\phi) = \frac{opposite}{hypotenuse}

⇒  tan(\phi) = \frac{324}{54}

⇒ (\phi) = tan^-^1(\frac{324}{54})

⇒ \phi=80.53

Nearest tenth it is 80.5 degrees.

So,

Men In Black must shoot the laser gun at an angle of 80.5 degrees.

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A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo
djverab [1.8K]

Answer:

a) y(t)=50000-49990e^{\frac{-2t}{25}}

b) 31690.7 g/L

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is \frac{dy}{dt}=R_{i}-R_{o}, where R_{i} is the rate of salt entering and R_{o} is the rate of salt going outside.

Then we have, R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}, and

R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}

So we obtain.  \frac{dy}{dt}=4000-\frac{2y}{25}, then

\frac{dy}{dt}+\frac{2y}{25}=4000, and using the integrating factor e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}, therefore  (\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}, we get   \frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}, after integrating both sides y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C, therefore y(t)= 50000 +Ce^{\frac{-2t}{25}}, to find C we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions y(0)=10, so 10= 50000+Ce^{\frac{-0*2}{25}}

10=50000+C\\C=10-50000=-49990

Finally we can write an expression for the amount of salt in the tank at any time t, it is y(t)=50000-49990e^{\frac{-2t}{25}}

b) The tank will overflow due Rin>Rout, at a rate of 80 L/min-40L/min=40L/min, due we have 500 L to overflow \frac{500L}{40L/min} =\frac{25}{2} min=t, so we can evualuate the expression of a) y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7, is the salt concentration when the tank overflows

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