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Elan Coil [88]
3 years ago
13

Which of the following lies in the ecliptic plane?

Physics
1 answer:
babymother [125]3 years ago
7 0
<h2>Answer: Earth's orbital path around the Sun</h2><h2></h2>

The <u>Ecliptic</u> refers to the orbit of the Earth around the Sun. Therefore, <u>for an observer on Earth it will be the apparent path of the Sun in the sky during the year, with respect to the "immobile background" of the other stars.</u>

<u />

It should be noted that the ecliptic plane (which is the same orbital plane of the Earth in its translation movement) is tilted with respect to the equator of the planet about 23\° approximately. This is due to the inclination of the Earth's axis.

Hence, the correct option is Earth's orbital path around the Sun.

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A car's gas tank contains 58.7 kg
Bogdan [553]

Answer:

721 kg/m^3

Explanation:

Trust me bro

8 0
3 years ago
A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acce
Delvig [45]

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

x=x_0+v_0t+at^2

Where

x_0 = Initial position

v_0 = Initial velocity

a = Acceleration

t= time

And the velocity can be expressed as,

v_f = v_0 + at

Where,

v_f = Final velocity

For our case we have that there is neither initial position nor initial velocity, then

x= at^2

With our values we have x = 401.4m, t=4.945s, rearranging to find a,

a=\frac{x}{t^2}

a = \frac{ 401.4}{4.945^2}

a = 16.41m/s^2

Therefore the final velocity would be

v_f = v_0 + at

v_f = 0 + (16.41)(4.945)

v_f = 81.14m/s

Therefore the final velocity is 81.14m/s

8 0
3 years ago
g An inductor used in a dc power supply has an inductance of 12.0 H and a resistance of It carries a current of 0.300 A. (a) Wha
valkas [14]

Answer:

Explanation:

Energy of an inductor = 1/2 L i²

L is inductance , i is current .

= 1/2 x 12 x .3²

= .54 J

4 0
3 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
What is the speed of a wave in (m/s) with a 5 meter wavelength and a period of 20 seconds?
arlik [135]

Answer: 0.25 m/s

Explanation:  Speed = wavelengt · frequency  

v = λf   and frequency is 1/period  f = 1/T

Then v = λ/T = 5 m / 20 s = 0.25 m/s

6 0
2 years ago
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