PHYSICS COLLEGE

Students run an experiment to determine the rotational inertia of a large spherically shaped object around its center. Through experimental data, the students determine that the mass of the object is distributed radially. They determine that the radius of the object as a function of its mass is given by the equation r = km², where k = 3. Which of the following is a correct expression for the rotational inertia of the object?

(A) m3
(B) 1.8 m3
(C) 3.6 m3
(D) 6 m3
(E) 9 m3

Answers

Answer 1

Answer:

Answer:

Explanation:

$r=km^2\n$ = $3m^2$

Since the object is a solid sphere, the equation for rotational inertia is:

$I = (2)/(5)mr^2$

$I=(2)/(5)m(3m^2)^2=(2)/(5)*9m^5=3.6m^5$

Answer 2

Answer:

Final answer:

The provided question seems to have a discrepancy as the calculated value of rotational inertia for a spherical object with a given mass-radius relationship is 4.5M³, which does not match any of the supplied answer choices.

Explanation:

The question is asking for the correct expression for the rotational inertia of a spherically shaped object with mass distribution given by the radius as a function of mass (r = km² where k = 3). The rotational inertia, or moment of inertia, for a solid sphere is given by the formula ⅒MR², where M is the mass of the sphere, and R is its radius. Considering that R is defined by r = km², we substitute R with km² in the formula:

I = ⅒M(km²)² = ⅒Mk²m⁴ = ⅒Mk²M²

Since k = 3, we further simplify the expression:

I = ⅒M(3M)² = ⅒(3²)M³ = ⅒ × 9M³ = 4.5M³

However, none of the options (A) to (E) match the value 4.5M³, which indicates there may be an error in the supplied options or an error within the initial assumptions or question parameters. It's important to recheck the given data and the calculation steps to ensure accuracy. If the question and the parameters are indeed accurate as stated, additional information or clarification would be necessary.

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C. Lenses refract light; mirrors do not.
D. Lenses focus light; mirrors do not.

Answers

This question involves the concepts of reflection and refraction.

The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".

LENSES AND MIRRORS

When it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the bending of light rays, while passing through a medium, without any rebound or absorption.

Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.

Learn more about reflection and refraction here:

brainly.com/question/3764651

Answer:

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What minimum distance would you have to hit a baseball from the center of the earth so that it would eventually reach the moon? Assume you can hit the ball directly along the line that connects the centers of the earth and moon. The distance between the centers of the earth and moon is ???? = 3.82 × 108 m.

Answers

Answer:

d = 3.44 x 10⁸ m

Explanation:

The minimum distance required will be the distance from the centre of the earth to a point where gravitational intensity due to both earth and moon becomes equal . Once this point is reached , moon will attract the baseball on its own .

Let this distance be d from the centre of the earth

So GM / d² = G m / ( 3.82 x 10⁸ - d )²

M is mass of the earth , m is mass of the moon

M / m = ( d / 3.82 x 10⁸ - d )²

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The record for the world’s loudest burp is 109.9 dB, measured at a distance of 2.5 m from the burper. Assuming that this sound was emitted as a spherical wave, what was the power emitted by the burper during his record burp?

Answers

Answer:

Power of the source is 7.7 W

Explanation:

As we know that the sound level is measured as

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now we have

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so we have

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Answers

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Answers

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Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31. They are issued at $508,050 when the market rate is 12%.1. Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225 $193,800
Par value at maturity 570,000
Total repaid 763,800
Less amount borrowed 645 669
Total bond interest expense $118.131
2. Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Period End Unamortized Discount Carrying Value
01/01/2019
06/30/2019
12/31/2019
06/30/2020
12/31/2020
3. Record the interest payment and amortization on June 30. Note:
Date General Journal Debit Credit
June 30
4. Record the interest payment and amortization on December 31.
Date General Journal Debit Credit
December 31

Answers

Answer:

1) Determine the total bond interest expense to be recognized.

Total bond interest expense over life of bonds:

Amount repaid:

8 payments of $24,225: $193,800

Par value at maturity: $570,000

Total repaid: $763800 (193,800 + 570,000)

Less amount borrowed: $508050

Total bond interest expense: $255750 (763800 - 508,050)

2)Prepare a straight-line amortization table for the bonds' first two years.

Semiannual Interest Period End; Unamortized Discount; Carrying Value

01/01/2019 61,950 508,050

06/30/2019 54,206 515,794

12/31/2019 46,462 523,538

06/30/2020 38,718 531,282

12/31/2020 30,974 539,026

3) Record the interest payment and amortization on June 30:

June 30 Bond interest expense, dr 31969

Discount on bonds payable, Cr (61950/8) 7743.75

Cash, Cr ( 570000*8.5%/2) 24225

4) Record the interest payment and amortization on December 31:

Dec 31 Bond interest expense, Dr 31969

Discount on bonds payable, Cr 7744

Cash, Cr 24225

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