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vladimir2022
11 days ago
5

The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i

s 1.5multiply1011 m. (a) Calculate the magnitude of the gravitational force exerted by the Sun on the Earth. N (b) Calculate the magnitude of the gravitational force exerted by the Earth on the Sun.
Physics
1 answer:
serg [2.5K]11 days ago
7 0
a) 3.56 x 10^22 N. b) 3.56 x 10^22 N. The sun’s mass is M = 2 x 10^30 kg, while the Earth's mass is m = 6 x 10^24 kg, with a distance of R = 1.5 x 10^11 m separating them. Applying Newton's law for gravitational force F = G (mM / R²), where G = 6.67 × 10^-11 m^3 kg^-1 s^-2 gives us F = 3.56 x 10^22 N. A) The gravitational force by the sun on Earth equates to the force exerted by Earth on the sun, which is also 3.56 x 10^22 N.
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Which statements describe properties of stars check all that apply
Keith_Richards [2263]

Answer:

Stars generate energy by the process of nuclear fusion.

They are large entities composed of gaseous elements.

The main constituents of stars are hydrogen and helium.

Explanation:

Stars are colossal objects with extensive gravitational forces causing them to contract, which allows fusion to take place: the atomic nuclei in the star's core are drawn very close together due to gravity and elevated temperatures, leading to the fusion reaction. This fusion serves as the energy output for a star.

Conversely, it is true that stars predominantly consist of hydrogen and helium (two hydrogen nuclei can fuse to become helium), which implies that a star is essentially an enormous ball of gas without a solid surface suitable for standing on.

As for the presence of water on a star, it is simply impossible. The extreme temperatures found in stars are far too high for water to exist in any liquid state on their surfaces.

7 0
3 days ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
kicyunya [2264]

Answer:

The particle's energy in its ground state is E₁=1.5 eV.

Explanation:

For a particle with mass m in the nth energy level of an infinite square well potential of width L , the energy E_{n} is given by:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1) and in the first excited state (n=2), where the energy is noted as E₂= 6.0 eV. Substituting into the above equation yields:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

Thus, we can express the ground state's energy as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Ultimately

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
23 hours ago
A magic medallion is suspended from a string inside a compartment of Hogwarts Express which is running straight westwards on hor
Yuliya22 [2438]

Answer:

a = 1 m/s²  and

Explanation:

The initial two sections are depicted in the attachment

Newton's second law is applied along each axis

Y axis

      Ty - W = 0        

      Ty = w

X axis

     Tx = m a

Using trigonometry, the tension components are determined

    Sin θ = Ty / T

    Ty = T sin θ

    Cos θ = Tx / T

    Tx = T cos θ

Acceleration calculation is carried out using kinematics

   Vf = Vo + a t

   a = (Vf -Vo) / t

   a = (20 -10) / 10

   a = 1 m/s²

Now we apply the substitution in Newton's equations

     

  T Sin θ = mg

  T cos θ = ma

Next, we divide both equations

  Tan θ = g / a

  θ = tan⁻¹ (g / a)

  θ = tan⁻¹ (9.8 / 1)

  θ = 84º

The equation for the angle does not depend on the mass, hence the angle remains unchanged

4 0
4 days ago
You are traveling upstream on a river at dusk. you see a buoy with the number 4 and a flashing red light. what should you do?
kicyunya [2264]
Traveling upstream, which means going against the current of the river, involves important navigation decisions. Buoys and markers indicate how to proceed safety. If you see a buoy with number 4 flashing red as you navigate at dusk, you should pass it on your starboard side (right side). Another buoy, for example, number 5, would have a different indication.
8 0
25 days ago
Starting from rest, the boy runs outward in the radial direction from the center of the platform with a constant acceleration of
Softa [2035]
The velocity when t=3 seconds is V=1.56 m/s. Given that acceleration a=0.5 m/s², angular velocity ω=0.2 rad/s, and time t=3 seconds, we calculate radial velocity Vr as Vr = a*t, leading to Vr = 1.5 m/s. The radius r is then calculated as r = a*t²/2, yielding r = 2.25m. The tangential velocity Vt is Vt = ω*r, resulting in Vt = 0.45 m/s. Finally, the total velocity V is computed as V = √(Vr² + Vt²), giving V = 1.56 m/s.
6 0
4 days ago
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