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Andrews [41]
1 year ago
11

Which of the following are dwarf planets? Check all that apply. Ceres Namaka Eris Charon Haumea Makemake Pluto

Physics
2 answers:
kicyunya [14]1 year ago
6 0
Ceres: Yes!
Namaka: No!
Eris: Yes!
Charon: No. (it's a satellite, and dwarf planet's can't be satellites!)
Haumea: Yes!
Makemake: Yes!
Pluto: Yes!

Glad To Help;)
lorasvet [3.4K]1 year ago
3 0

Answer:

Dude above is correct

Explanation:

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The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin
KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

given i = 180 , Imax = 200 , f = 50  , t = ?

Put the give values in the equation above

180 = 200 sin 2πft

sin 2πft = .9

sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

360 x 50 t = 360n + 64

360 x 50 t =  64 ,  ( putting n = 0 for least value of t )

18000 t = 64

t = 3.55 ms  .

8 0
2 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
2 years ago
Driving your Ferrari through the Italian countryside at a speedy 88 m/s, you approach an opera diva singing a high C (1,046 Hz).
MrRissso [65]

Answer:

You will hear the note E₆

Explanation:

We know that:

Your speed = 88m/s

Original frequency = 1,046 Hz

Sound speed = 340 m/s

The Doppler effect says that:

f' = \frac{v \pm v0 }{v \mp vs}*f

Where:

f = original frequency

f' = new frequency

v = velocity of the sound wave

v0 = your velocity

vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.

Replacing the values that we know in the equation we have:

f' = \frac{340 m/s + 88m/s}{340 m/s} *1,046 Hz = 1,316.73 Hz

This frequency is close to the note E₆ (1,318.5 Hz)

7 0
1 year ago
A compressed spring has 16.2 J of elastic potential energy when it is compressed 0.30 m. What is the spring constant of the spri
nikdorinn [45]
Elastic potential = 1/2 x constant x square of compression lenght
So it's 360 N/m
7 0
2 years ago
Read 2 more answers
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

                                                                                  = 3.1536 x 10⁷ +0.840

                                                                                 = 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

                                                                               = 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

                                                                                   = 86400.0023 s

The time period of rotation is given by the formula,

                                <em>Tₐ = 2π /ωₐ</em>

                                 ωₐ = 2π / Tₐ

Substituting the values,

                                  ωₐ = 2π /  365.046306        

                                      = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

                                                                                    = 86400 s

Time period of rotation,

                                   Tₓ = 2π /ωₓ

                                    ωₓ = 2π / T

                                           =  2π /86400

                                          = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration

                                   α  = (ωₓ -   ωₐ) /  T₁

             = (7.272205217  x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

                                    α  = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

3 0
2 years ago
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