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choli [55]
2 years ago
11

Which phrases describe all the outer planets motion? Select two options

Physics
1 answer:
11111nata11111 [884]2 years ago
4 0

Answer:

slow revolution and  fast rotation

Explanation:

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Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th
melomori [17]

The units of G must be C. m³ / ( kg s² )

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem !

To find unit of Gravitational Constant can be carried out in the following way:

F = G \frac{m_1 ~ m_2}{R^2}

{[N]}= G\frac{{[kg]}{[kg]}}{{[m^2]}}

{[kg ~ m / s^2]}= G \frac{{[kg^2]}}{{[m^2]}}

G = \frac{{[kg ~ m / s^2]}{[m^2]}} {{[kg^2]} }

G = \frac{{[kg ~ m^3 / s^2]}} {{[kg^2]} }

G = \frac{{[m^3 / s^2]}} {{[kg]} }

\boxed {G = \frac{{[m^3]}} {{[kg ~ s^2]} }}

The unit of G must be \large {\boxed {\frac{m^3} {kg ~ s^2 }}}

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

5 0
3 years ago
Read 2 more answers
1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as
telo118 [61]

Explanation:

The tangential speed of Andrea is given by :

v=r\omega

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v

New angular speed is twice that of the Chuck's speed.

8 0
3 years ago
Research on the Big Five has indicated that approximately _____ percent of traits appear to be inherited.
Evgen [1.6K]
 D 50
Hopefully this helps!
 
4 0
2 years ago
Read 2 more answers
A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0
2 years ago
A beam of white light enters a prism of crown glass (n = 1.5) from air (n = 1.00). Once inside, the colors in the light disperse
eduard

Answer:

= 2.33

Explanation:

.According to snell's law:

n1sin i = n2sin r ,

where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,

i is angle of incidence,

r is angle of refraction.

Given that,

n1 = 1,

i = 51 degrees,

r = 19.5 degrees. ,

n2= ?

So,

1*sin 51 = n2 sin 19.5  

=> n2 = sin51 / sin19.5

 = 2.33

7 0
3 years ago
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