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anzhelika [568]
3 years ago
12

What is a small body that follows a highly elliptical orbit around the sun

Physics
2 answers:
Marianna [84]3 years ago
7 0
A comet is a small body of ice, rock, and dust that follows a highly elliptical orbit around the sun.
melomori [17]3 years ago
4 0
A Planet, such as (pluto)
You might be interested in
A ray of red light in air is incident at an angle of 30. on a
Vlad [161]

Answer:

20 degrees.

Explanation:

From Snell’s law of refraction:

sinθ1•n1 = sinθ2•n2

where θ1 is the incidence angle, θ2 is the refraction angle, n1 is the refraction index of light in medium1, and n2 is the refraction index for virgin olive oil. The incidence angle of the red light is θ1 = 30 degrees.

The red light is in air as medium1, so n1 (air) = 1.00029

So, to find θ2, the refracted angle:

sinθ1•1.00029 = sinθ2•1.464

sin(30)•1.00029 / 1.464 = sinθ2

0.5•1.00029 / 1.464 = sinθ2

sinθ2 = 0.3416291

θ2 = arcsin(0.3416291)

θ2 = 19.976 degrees

To the nearest degree,

θ2 = 20 degrees.

8 0
3 years ago
MULTIPLE CHOICE
Juli2301 [7.4K]

Answer:

answers d

Explanation:

hopes its healp

8 0
2 years ago
Read 2 more answers
How much heat in joules is required to heat a 50 g sample of aluminum from 71 ∘f to 142 ∘f?
Vsevolod [243]
A 15.75-g<span> piece of iron absorbs 1086.75 </span>joules<span> of </span>heat<span> energy, and its ... </span>How many joules<span> of </span>heat<span> are </span>needed<span> to raise the temperature of 10.0 </span>g<span> of </span>aluminum<span> from 22°C to 55°C, if the specific </span>heat<span> of </span>aluminum<span> is o.90 J/</span>g<span>”C2 .</span>
3 0
3 years ago
A television camera lens has a 17-cm focal length and a lens diameter of 6.0 cm. what is its number?
IRINA_888 [86]

Answer:

= 2.83

Explanation:

F number (N) is given by the formula;

  F- number = f/D

where f = focal length of lens and D = diameter of the aperture  

Therefore;

F number = 17 cm/6 cm

                <u> = 2.83</u>

3 0
3 years ago
A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane
Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, \phi=8.6\times 10^{-3}\ T/m^2

The magnetic flux linked through the loop is :

\phi=B.A

\phi=BA\ cos\theta

Here, \theta=0

B=\dfrac{\phi}{A}

or

B=\dfrac{\phi}{\pi r^2}

B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0
3 years ago
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