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maria [59]
3 years ago
7

A particular spiral galaxy can be approximated by a thin disk-like volume 62 Thousand Light Years in radius and 7 Hundred Light

Years thick. If this Galaxy contains 1,078 Billion stars, estimate the average distance between the stars in this galaxy. Hint: calculate the average volume per star in cubic Light Years, and then estimate the approximate linear dimension across such a volume. (Indicate your answer to one decimal place.)
Physics
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

Answer:

Approximate linear dimension is 2 light years.

Explanation:

Radius of the spiral galaxy r = 62000 LY

Thickness of the galaxy h = 700 LY

Volume of the galaxy = πr²h

                                   = (3.14)(62000)²(700)

                                   = (3.14)(62)²(7)(10)⁸

                                   = 84568×10⁸

                                   = 8.45\times 10^{12} (LY)³

Since galaxy contains number of stars = 1078 billion stars ≈ 1.078\times 10^{12}

Now volume covered by each star of the galaxy = \frac{\text{Total volume of the galaxy}}{\text{Number of stars}}

= \frac{8.45\times 10^{12} }{1.078\times 10^{12}}

= 7.839 Light Years

Now the linear dimension across the volume

= (\text{Average volume per star})^{\frac{1}{3}}

= (7.839)^{\frac{1}{3}}

= 1.99 LY

≈ 2 Light Years

Therefore, approximate linear dimension is 2 light years.

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Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the
mash [69]

Answer:

The value is }  N  =  66 \  blocks

Explanation:

From the question we are told that

The weight of the block is W_b  = 100 \  N

The dimension of the block is d =  0.400 m  \ X  \ 0.250 \  m  \  X  \ 0.130 \ m

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P_{2atm} =  2 *  1.013 *10^{5} =  202600 \  N/m^2

Generally 1 atm = 1.013 *10^{5} N/m^2

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

So

A =  0.250 *  0.130

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Generally the force due to this blocks is mathematically represented as

F =  N  *  W_b

Here N is the number of blocks

So

}  202600 =  \frac{N  *  100 }{ 0.0325}

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Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
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Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • mass of the larger disk = M_2\ =\ 1.6\ kg
  • Radius of the larger disk =R_2\ =\ 5.0\ cm\ =\ 0.05\ m
  • mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding,\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

Net torque on the smaller disk,

\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)

From eqn (1) and (2), we get,

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let 'a_2' be the acceleration of the block in the second case,

From the above expression,

\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2

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3 years ago
(c) An object of mass 100 kg is accelerated uniformly from a velocity of 15m/s to 20 m/s in 10 s.
Pavlova-9 [17]

Explanation:

momentum = mass x velocity

initial momentum = 100 x 15 = 1500kgm/s

after momentum = 100 x 20 = 2000kgm/s

a =(v-u)/t

a = (20-15)/10

a = 5/10

a = 0.5m/s²

f = ma

f = 100 x 0.5

f = 50N

7 0
4 years ago
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