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Ira Lisetskai [31]
3 years ago
8

Suppose you see two main-sequence stars of the exact same spectral type. Star 1 is dimmer in apparent brightness than Star 2 by

a factor of 100. What can you conclude? (Neglect any effects that might be caused by interstellar dust and gas.)
Physics
1 answer:
katrin2010 [14]3 years ago
3 0

Options:

A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.

B. Star 1 is 100 times more distant than Star 2.

C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.

D. Star 1 is 10 times more distant than Star 2.

E. Star 1 is 100 times nearer than Star 2.

Answer:

D. Star 1 is 10 times more distant than star 2

Explanation:

For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.

Luminosity, L = 1/r²

Where r is the distance of the star to the earth

Since star 1 is dimmer in brightness than star 2 by a factor of 100,

L₁/L₂ = 1/100

i.e. L₁ = 1, L₂=100

L₁ = 1/r₁² ............(1)

1 =  1/r₁²

L₂ = 1/r₂²

100 =  1/r₂² .........(2)

divide equation (2) by equation (1)

100/1 = ( 1/r₂² )/ (1/r₁²)

100 = (r₁/r₂)²

r₁/r₂ = √100

r₁/r₂ = 10

r₁ = 10r₂

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Car A is traveling at 20.0 m/s and car B at 27.0 m/s.
Savatey [412]

Take the moment car A starts to accelerate to be the origin. Then car A has position at time <em>t</em>

<em>x</em> = (20.0 m/s) <em>t</em> + 1/2 (2.10 m/s²) <em>t</em>²

and car B's position is given by

<em>x</em> = 300 m + (27.0 m/s) <em>t</em>

<em />

Car A overtakes car B at the moment their positions are equal:

(20.0 m/s) <em>t</em> + 1/2 (2.10 m/s²) <em>t</em>² = 300 m + (27.0 m/s) <em>t</em>

300 m + (7.00 m/s) <em>t</em> - (1.05 m/s²) <em>t</em>² = 0

==>  <em>t</em> ≈ 20.6 s

4 0
3 years ago
A certain dam generates 120 MJ of mechanical (hydroelectric) energy each minute. If the conversion from mechanical to electrical
Karolina [17]

Answer:

electric energy ( power ) = 300000 W

Explanation:

given data

mechanical (hydroelectric) energy = 120 MJ/min = 2000000 J/s

efficiency = 15 % = 0.15

solution

we know that Efficiency of electric engine is expression  as

Efficiency = Mechanical energy ÷ electric energy  ......................1

and here dam electrical power output is

put here value in equation 1

electric energy ( power ) = Efficiency × Mechanical energy ( power )

electric energy ( power ) = 0.15 × 2000000 J/s

electric energy ( power ) = 300000 W

8 0
3 years ago
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.8×104 m/s when at a distance
natali 33 [55]

Answer:

9.4 x 10⁴ m/s

Explanation:

In circular or elliptical orbits , the angular momentum is conserved because torque is zero.  

mr₁v₁ = mr₂v₂

2.2 x 10¹¹ x 1.8 x 10⁴ = 4.2 x 10¹⁰ v₂

v₂ = 9.42 x 10⁴ m/s.

= 9.4 x 10⁴ m/s

7 0
3 years ago
A and B start walking in the same direction at the same time around a circular park of diameter 4200 m. If A walks 10 m more tha
Liono4ka [1.6K]

Answer:

Va = 5000 m / 3600 s = 1.39 m/s

(Va - Vb) 60 = 10

Vb = Va - .167 = 1.22 m/s

(Va - Vb) T = 4200 Π      where T is time for A to complete 1 more lap

.17 T = 4200 Π

T = 24700 Π  time for A to again catch B

N = 1.39 * 24700 Answer:

Va = 5000 m / 3600 s = 1.39 m/s

(Va - Vb) 60 = 10

Vb = Va - .167 = 1.22 m/s

(Va - Vb) T = 4200 Π      where T is time for A to complete 1 more lap

.17 T = 4200 Π

T = 24700 Π  time for A to again catch B

N = 1.39 * 24700 Π / (4200 Π) = 8.2   laps

A will make 8 but not 9 rounds before catching B

6 0
2 years ago
A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal
Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

So, at half of the maximum height,

h = \frac{13.7}{2}=6.9 m

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s

And so, the speed of the stone at half of the maximum height is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

v_1 = 18.9 m/s

while the speed at half of the maximum height (part b) is

v_2 = 22.2 m/s

So the difference is

\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0
4 years ago
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