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lorasvet [3.4K]
3 years ago
9

Black vultures excel at gliding flight; they can move long distances through the air without flapping their wings while undergoi

ng only a modest drop in height. A vulture in a typical glide in still air moves along a path tipped 3.5∘ below the horizontal. If the vulture moves a horizontal distance of 100 m, how much height does it lose?
Physics
1 answer:
GenaCL600 [577]3 years ago
4 0

Answer:

The vulture loses 6.1 m height

Explanation:

Please see the attached figure.

The horizontal distance and the loss of height form a 90º triangle.

The loss of height is  the side opposite the given angle (3.5º) and the 100 m horizontal distance is adjacent the angle.

Then, using trigonometric rules:

(1)   sin 3.5º = h / hyp

(2)  cos 3.5º = distance / hyp

where

h = height lost during the flight.

hyp = hypotenuse of the triangle

Using (2) we can calculate the hypotenuse:

cos 3.5º = 100 m / hyp

hyp = 100 m / cos 3.5º = 100.2 m

with the hypotenuse we can now calculate the loss of height using (1):

sin 3.5º = h / hyp

sin 3.5º = h / 100.2 m

sin 3.5º * 100.2 m = h

<u>h = 6.1 m</u>

( very modest drop in height indeed!)

Download pdf
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A box sits on the back of a flatbed truck. If the coefficient of static friction between the box and the truck bed is μs = 0.400
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Answer:

28,699m

Explanation:

The force to make the box move should be <u>μs.N=μs.m.g=m.</u><u>|</u><u>a</u><u>|</u>

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and a the acceleration of the truck.

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as the truck is stopping, the acceleration is negative. then,

x = v \times t  -  \frac{1}{2} \times  |a| \times   {t}^{2}

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x = v \times (v \div   |a| ) -  \frac{1}{2} \times  |a| \times   {(v \div   |a|)}^{2}

x = v \times (v \div  μs.g ) -  \frac{1}{2} \times  |a| \times   {(v \div   μs.g)}^{2}

x =   \frac{{v}^{2}}{μs \times g} -  \frac{1}{2} \times  \frac{{v}^{2}}{μs \times g} \\ x = \frac{1}{2} \times  \frac{{v}^{2}}{μs \times g} \\ x = 0.5 \times  \frac{{(15m/s) }^{2} }{0.4 \times 9.8m/ {s}^{2} }  = 28.699m

28,699m

7 0
2 years ago
What is the quantity of heat energy required to convert 10g cube of ice at -30oC to steam at 120oC. also draw a graph of tempera
julia-pushkina [17]

30,869.2 J

Explanation:

Given parameters:

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Final temperature = 120°C

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Specific heat capacity of steam  = 1.996J/g°C

Latent heat of fusion of water(l) = 334J/g

Latent heat of vaporization = 2230J/g

Unknown:

Quantity of heat required = ?

Temperature-energy graph = ?

Solution:

The temperature energy profile is attached to this solution.

q = mc∅ₓ + mlₓ + mc∅ₙ + mlₙ + mc∅ₐ

qₓ = mc∅ₓ in converting ice from -30°C to ice at 0°C

qₓ is the amount of heat supplied to the ice that changes its temperature from -30°C to that at freezing point:

qₓ  = 10 x 4.2 x (0-(-30)) = 10 x 2.1 x 30 = 630J

qₓ = mlₓ in converting ice to water

qₓ here is the latent heat used to break the ice bonds without a change in temperature:

qₓ = ml = 10 x 334 = 3340J

qₙ = mc∅ₙ is the heat from water at 0°C to boiling point.

This is the heat required to take water from freezing temperature to its boiling point

qₙ = mc∅ₙ  = 10 x 4.2 x (100 - 0) = 4200J

qₙ  = mlₙ is the heat in vaporizing water

In vaporizing water, there is no temperature change when the hydrogen bonds are broken. The heat supplied is not used to raise temperature. We use the latent heat of vaporization:

qₙ   = 10 x 2230 = 22300J

qₐ = mc∅ₐ from vapor at 100°C to 120°C

This is the heat used to raise the temperature of vapor:

qₐ = 10 x 1.996 x (120-100) = 399.2J

The total heat:

  q = qₓ + qₙ + qₐ = 630J + 3340J + 4200J + 22300J + 399.2J =30,869.2 J

Learn more:

Specific heat brainly.com/question/7210400

#learnwithBrainly

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