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Ira Lisetskai [31]

1 year 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.)

1 answer:

katrin2010 [14]1 year 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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schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

7 0

1 year ago

Table 2.4 shows how the dispacement of a runner changed during a sprint race. Draw a dispacement-time graph to show this data, a

GalinKa [24]

4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12

8 0

1 year ago

While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$

$m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sin$ Ф

60 ·8 + 0 = (60 + 80) $V_{f}sin$ Ф

480 = 140 $V_{f} sin$ Ф................. (I)

y-axis component form (+y north);

$P_{Mix}$ + $p_{Pix}$ + $j_{x}$ = $P_{Mfx}$ + $P_{pfx}$

$m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos$ Ф

0 + 80.9 = (60 + 80) $V_{f}cos$ Ф

720= 140 $V_{f}cos$ Ф

140Vf= $\frac{720}{cos}$ Ф......................................(2)

Substituting (2) into (1) to give the angle;

480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

7 0

2 years ago

Ocean waves are observed to travel to the right along the water surface during a developing storm. A Coast Guard weather station

Nuetrik [128]

Answer:

The amplitude is 2.3 m

The Wavelength is 8.6 m

The frequency is 0.16 Hz

The time period is 6.25 sec

The equation that governs the behavior is $Y=(2.3)sin[(\frac{2\pi}{8.6} )x -(\frac{2\pi}{6.2} )t]$

Explanation:

The explanation is shown on the first uploaded image

6 0

2 years ago

A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu

Pepsi [2]

Answer:

$v = 10.89\ m/s$

Explanation:

given,

radius of loop = 12.1 m

to find the minimum speed transverse by the rider to not to fall out upside down

centripetal force = $\dfrac{mv^2}{r}$

gravitational force = m g

computing both the equation]

$mg = \dfrac{mv^2}{r}$

$v = \sqrt{rg}$

$v = \sqrt{12.1 \times 9.8}$

$v = \sqrt{118.58}$

$v = 10.89\ m/s$

5 0

2 years ago