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kozerog [31]
3 years ago
7

slader A jet is circling an airport control tower at a distance of 15.9 km. An observer in the tower watches the jet cross in fr

ont of the moon. As seen from the tower, the moon subtends an angle of 8.74x10-3 radians. Find the distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.
Physics
1 answer:
liubo4ka [24]3 years ago
4 0

Answer:

y = 138.96 m

Explanation:

The angle subtended by the moon is the mean of the angle of the arc between the two most extreme points of the moon, we can see that the angle is very small, so we can approximate this arc to a straight line and then use the trigonometric relationships

         sin θ = y / L

where L = 15.9 10³ m and θ = 8.74 10⁻³ rad

          y = L sin θ

          y = 15.9 10³ sin (8.74 10⁻³)

         y = 15.9 10³    0.0087399

         y = 138.96 m

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Hello there! Your answer is (255/25=9) hope it helps and have a wonderful day!
6 0
4 years ago
Two moles of helium gas initially at 438 K and 0.44 atm are compressed isothermally to 1.61 atm. Find the final volume of the ga
docker41 [41]

Answer:

44.64335 L

Explanation:

R = Gas constant = 8.314 J/mol K = 0.08205 L atm/mol K

P = Pressure

V = Volume

T = Temperature = 438 K

1 denotes initial

2 denotes final

From ideal gas law we have

PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2\times 0.08205\times 438}{0.44}\\\Rightarrow V=163.35409\ L

So,

P_1V_1=P_2V_2\\\Rightarrow V_2=\dfrac{P_1V_1}{P_2}\\\Rightarrow V_2=\dfrac{0.44\times 163.35409}{1.61}\\\Rightarrow V_2=44.64335\ L

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5 0
3 years ago
In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th
lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, \theta=20.0°

Mass of package is, m=2.50\ kg

Initial speed of package is, u=2.00\ m/s

Final speed of the package at the bottom is, v=0\ m/s

Distance of travel along the incline is, d=12.0\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Let the coefficient of kinetic friction be \mu.

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

a=\mu g\cos\theta-g\sin\theta ----------------- 1

Now, using equation of motion, we have:

v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)

Solving for 'a', we get:

-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2

Now, plug in the value of 'a' in equation (1). This gives,

\mu g\cos\theta-g\sin\theta=\frac{1}{6} ( Neglecting negative sign)

Plug in all the given values and solve for \mu. This gives,

9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382

Therefore, the coefficient of kinetic friction is 0.382.

5 0
4 years ago
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Answer:

what is your exact question.

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Answer:

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