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Ira Lisetskai [31]

2 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.)

1 answer:

katrin2010 [14]2 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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A 4.0 Ω resistor, an 8.0 Ω resistor, and a 12.0 Ω resistor are connected in parallel across a 24.0 V battery. What is the equiva

dsp73

PART A)

Equivalent resistance in parallel is given as

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

now we have

$\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12}$

$R = 2.18 ohm$

PART B)

since potential difference across all resistance will remain same as all are in parallel

so here we can use ohm's law

$V = iR$

for 4 ohm resistance we have

$24 = 4(i)$

$i = 6 A$

PART C)

since potential difference across all resistance will remain same as all are in parallel

so here we can use ohm's law

$V = iR$

for 8 ohm resistance we have

$24 = 8(i)$

$i = 3 A$

3 0

2 years ago

Read 2 more answers

A heat recovery system (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loop

Annette [7]

Answer: Volume of Freon = 12.23. $10^{3}$ L

Explanation: First, the temperatures have to be in the same unit, so all the tmeperatures is transformed into Kelvin:

T₁ = 192K

T₂ = $\frac{5}{9}$ . (F - 32)+273.15

T₂ = $\frac{5}{9}$ . (- 65 - 32)+273.15

T₂ = 219.26K.

Second, BTU is an unit of energy. It can be transformed into Joules by the relation 1BTU = 1055.06J.

Q = 1055.06 . 46.9 = 49482.314J

The letter Q represents Heat and is calculated as Q = m.Cp.ΔT, where:

m is mass of the element;

Cp is the heat capacity specific for each element;

ΔT is the difference between the final and initial temperature;

Third, it will be needed the properties of the Freon:

Freon (CF2Cl2) = 120.91g/mol; Cp = 74J/mol.K; ρ = 1.49kg/m³

Fourth, we calculate the mass of Freon necessary for the remove of 46.9BTU of energy from the system:

Q = m.Cp.ΔT

m= $\frac{Q}{c(T-T_{0} )}$

m = $\frac{49482.314}{74(219.26-192)}$

m = 18228.214g or m=18.228Kg

Fifth, using the density of Freon, Volume can be found

ρ = $\frac{m}{V}$

V = m / ρ

V = 18.228 / 1.49

V = 12.23m³

As SI for density is Kg/m³, the volume found is in m³.

Cubic meters is related to Liters as the following: 1m³=1000l.

So, volume of Freon = 12.23. $10^{3}$ L

The volume of Freon needed is 12.23. $10^{3}$ L.

7 0

2 years ago

The temperature, T, of a gas is jointly proportional to the pressure P of the gas and the volume V occupied by the gas. Use C as

AnnZ [28]

Answer:

T=C*P*V

Explanation:

It is said that a variable - let's call 'y' -, is proportional to another - let's call it 'x' - if x and y are multiplicatively connected to a constant 'C'. It means that their product (x*y) can be always equaled to the constant 'C' or their division ( $\frac{x}{y}$ ) can be always equaled to 'C'. The first case is the case of the inverse proportionality: It is said that x and y are inversely proportional if

$x*y=C$

The second case is the case of the direct proportionality: It is said that x and y are directly proportional if

$\frac{x}{y} =C$ : x is directly proportional to y.

or

$\frac{y}{x} =C$ : y is directly proportional to x.

Always that any text does not specify about directly or inversely proportionality, it is assumed to mean directly automatically.

For our case, we are said that the temperature T is proportional to the pressure P and the volume V (we assume that it means directly); it is a double proportionality but follows the same rules:

If T were just proportional to P, we would have:

$\frac{T}{P} =C$

If T were just proportional to V, we would have:

$\frac{T}{V} =C$

As T is proportional to both P and V, the right equation is:

$\frac{T}{P*V}=C$

In order to isolate the temperature, let's multiply (P*V) at each side of the equation:

$\frac{T}{P*V}*(P*V)=C*(P*V)\\T=C*P*V$

3 0

2 years ago

At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m

Margaret [11]

Answer:

Part a)

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = 1.06 s$

Part c)

$L = 4.86 m$

Explanation:

Part a)

The height of the diving board is given as

$h = 5.5 m$

now the speed of the diver is given as

$v_0 = 2.7 m/s$

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

$y = v_y t + \frac{1}{2}at^2$

$h = 0 + \frac{1}{2}gt^2$

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = \sqrt{\frac{2h}{g}}$

plug in the values in the above equation

$t = \sqrt{\frac{2(5.5 m)}{9.81}$

$t = 1.06 s$

Part c)

Horizontal distance moved by the diver is given as

$d = v_0 t$

$d = 2.7 \times 1.06$

$d = 2.86 m$

so the distance from the edge of the pool is given as

$L = 2.86 + 2$

$L = 4.86 m$

4 0

2 years ago

A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.

Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

$v^{2}=u^{2}+2as$

$0^{2}=35^{2}- 2 \times 9.8 \times h$

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

8 0

2 years ago