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2 years ago

4.0km at 25 west of south, then 2.0km at 15 north east

Physics

1 answer:

Aliun [14]2 years ago

3 0

please, give the question properly.

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joja [24]

D.
Always use the right tool to get accurate measurements

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3 years ago

Two airplanes leave an airport at the same time. The velocity of the first airplane is 750 m/h at a heading of 51.3°. The veloci

abruzzese [7]

Refer to the diagram shown below.

In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.

The second airplane travels
b = 620*2.4 = 1488 mile

The angle between the two airplanes is
163° - 51.3° = 111.7°

Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
= 3.24 x 10⁶ + 2.2141 x 10⁶
c = 2335.41 miles

Answer: 2335.4 miles

4 0

2 years ago

Read 2 more answers

The critical angle for water is 49°. If a ray of light

Sonja [21]

Answer:

Snell's Law states

Ni sin i = Nr sin r

Judging from the question the source of the ray is in the water (directed up)

or NI = 1 / sin 49 Ni = 1.325 deg the critical angle

From inside the pond:

Nr = 1.325 * sin 45 / 1 = 94 deg

So refraction can occur outside the pond and you do not have total internal refection.

3 0

2 years ago

Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge b

Elena L [17]

Answer:

a) 0.167 μC/m^2

b) 1.887 * 10^4 V/m

Explanation:

Hello!

First let's find the surface charge density:

Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:

σ = Q/A

Where A is the area of the satellite, which is:

A=4πr^2 = πd^2 = π(1.9m)^2

Therefore:

σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2

Now let's calculate the electric field

Just outside the surface of the satellite the elctric field will be:

E = σ/ε0

Where ε0=8.85×10^−12 C/Vm

Therefore:

E = (0.167*10^-6 C/m^2) / (8.85*10^-12 C/Vm) = 0.01887 *10^6 V/m

E = 1.887 * 10^4 V/m

5 0

2 years ago

When you observe a distant galaxy whose photons have traveled for 10 billion years before reaching earth we are seeing that gal?

expeople1 [14]

In reality we don't see the galaxy we see it's reflection .. the light hits or got emitted by the star travel all the way long to hit our eyes .. we see their reflection . everything around you that you see is it's reflection

8 0

3 years ago