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Studentka2010
18 days ago
9

Suppose that now you want to make a scale model of the solar system using the same ball bearing to represent the sun. How far fr

om it would you place a sphere representing the earth? (Center to center distance please.) The distance from the center of the sun to the center of the earth is 1.496×10111.496×1011 m and the radius of the sun is 6.96×1086.96×108 m.
Physics
1 answer:
serg [2.5K]18 days ago
7 0

Response:

d = 0.645 m (assuming the ball bearing's radius is 3 mm)

Clarification:

Here's what we know:

  • The distance from the sun's center to earth's center is 1.496x10¹¹m = d_{e}
  • The sun's radius measures 6.96x10⁸m = r_{s}

We need to estimate a radius for the ball bearing; let's take the radius to be 3 mm = r_{b}.

First, we need to determine how much larger the sun's radius is compared to the ball bearing's radius using the equation below:

\frac{r_{s}}{r_{b}} = \frac{6.96\cdot 10^{8}m}{3\cdot 10^{-3}m} = 2.32\cdot 10^{11}

Now, let's figure out the distance from the sun's center to the center of a spherical model for the earth, d_{s}:

[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m

I trust this is beneficial for you!

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What is meant in astronomy by the phrase "adaptive optics?
kicyunya [2264]

Response: a. The mirrors and eyepiece of a large telescope are designed with spring-loaded components to quickly return to a predetermined position.

Justification:

Adaptive optics refers to a technique employed by various astronomical observatories to compensate in real-time for the atmospheric turbulence that impacts astronomical imaging.

This is executed by integrating advanced deformable mirrors into the telescope's optical pathway, operated by a set of computer-controlled actuators. This allows for obtaining clearer images despite the atmospheric fluctuations that create distortions.

It is crucial to note that this process requires a moderately bright reference star located closely to the object being studied.

However, locating such stars is not always feasible, prompting the use of a strong laser beam directed at the upper atmosphere to create artificial stars.

7 0
4 days ago
During a snowball fight two balls with masses of 0.4 and 0.6 kg, respectively, are thrown in such a manner that they meet head-o
Yuliya22 [2420]

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

Speed of 1 Kg mass post-collision =?

Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

8 0
20 days ago
(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe
serg [2593]

Answer:

(a) the coefficient of friction is 0.451

This was derived using the energy conservation principle (the total energy in a closed system remains constant).

(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.

Explanation:

For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.

Thank you for your attention; I trust this is beneficial to you.

4 0
20 days ago
A beaker contain 200mL of water<br> What is its volume in cm3 and m3
Sav [2226]
The volumes are 200cm3 and 0.0002m3
7 0
14 days ago
Read 2 more answers
A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air
Sav [2226]

Answer:

Height (h) = 17 m

Velocity (v) = 18.6 m/s

Explanation: This problem can be solved using kinematic motion equations.

Given Data

Initial velocity (u) = 0

Acceleration (a) = g

Time (t) = 1.9 seconds

First, we calculate the height.

s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m

Then, we find the final velocity

v=u+at\\v=0+9.81*1.9\\v=18.639

The acceleration graph is a linear representation described by y=9.8, as it remains constant:

The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):

The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):

These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.

5 0
1 month ago
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