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vladimir1956 [14]
3 years ago
14

The North Star is located in the Big Dipper. True or False

Physics
2 answers:
harina [27]3 years ago
4 0

Answer:

False

Explanation:

It is located in the little dipper whose stars are more faint.

Maurinko [17]3 years ago
4 0
False, the North Star is located at the end of the handle of the Little Dipper, as opposed to the Big Dipper. Hope this helps!
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I don't understand this question at all, can I please get some help?
beks73 [17]
V^2/R=180W
v=root 180R
4 0
3 years ago
An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are inv
PolarNik [594]

Complete Question

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick

Answer:

The  largest total area of the oil slick  A = 8.257 *10^{9} \ m^2

Explanation:

From the question we are told that

     The volume of oil the escaped is  V  = 18000 \ L

    The refractive index of oil is n_o = 1.1

     The refractive index of water is n_w = 1.33

      The wavelength of the light  is \lambda = 485 \ nm =  485 * 10^{-9} \ m

         

Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

          d = m *\frac{\lambda}{2n_w}

Where is the order of interference of the light and it value ranges from 1, 2, 3,...n

It is usually take as 1 unless stated otherwise by the question

substituting value

      d = 1 * \frac{485 *10^{-9}}{2 * 1.1}    

      d = 218 nm    

The are can be mathematically evaluated as

        A = \frac{V}{d}

Substituting values

        A = \frac{18000}{218*10^{-8}}

        A = 8.257 *10^{9} \ m^2

6 0
2 years ago
A ball is thrown straight upward and rises to a maximum
Leviafan [203]

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

v_f^2 - v_i^2 = 2 a d

here we know that

v_f = 0

also we have

a = -9.81 m/s^2

so we have

0 - v_i^2 = 2(-9.81)(16)

v_i = 17.72 m/s

Now we need to find the height where its speed becomes half of initial value

so we have

v_f = 0.5 v_i

now we have

v_f^2 - v_i^2 = 2 a d

(0.5v_i)^2 - v_i^2 = 2(-9.81)h

-0.75v_i^2 = -19.62 h

0.75(17.72)^2 = 19.62 h

h = 12 m

3 0
3 years ago
Write fulk form of FPS?<br>​
zzz [600]

Answer:

<u>Foot per second. Foot-pound-second system. Frames per second, the frequency (rate) at which consecutive images (frames) appear on a display.</u>

Explanation:

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4 0
3 years ago
Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the cha
leva [86]

Answer:

Explanation:

a ) Between r = 0 and r = r₁

Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .

b ) From r = r₁ to r = r₂

At distance r , charge contained in the sphere of radius r

volume charge density x 4/3 π r³

q = Q x r³ / R³

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q x r³ / ε₀R³

E= Q x r / (4πε₀R³)

E ∝ r .

c )

Outside of r = r₂

charge contained in the sphere of radius r = Q

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q  / ε₀

E = Q  / 4πε₀r²

E ∝ 1 / r² .

6 0
3 years ago
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