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3 years ago

The temperature of a star is 6000K and its luminosity is 1000. Which type of star is it?

Physics

2 answers:

Nesterboy [21]3 years ago

7 0

Answer:it’s a supergiant

Explanation: if this is what you are doing for your mastery test then this is the correct answer

Annette [7]3 years ago

6 0

Answer:

F-type star

Explanation:

Stars are classified on the basis of temperature and luminosity in the H-R diagram.

Based on the temperature, the stars classified as:

O, B, A, F, G, K, and M.

O-type stars are the hottest stars and M-type stars are the coolest ones.

A star having temperature 6000 K and luminosity 1000 would be a F-type star.

Luminosity depends on temperature and size of a star. More the temperature, more would be its luminosity.

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A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take t

marin [14]

Answer:

$T=66262.4s$

Explanation:

From the question we are told that:

Altitude $A=2.90 *10^7$

Mass $m=5.97 * 10^{24} kg$

Radius $r=6.38 *10^6 m.$

Generally the equation for Satellite Speed is mathematically given by

$V=(\frac{GM}{d} )^{0.5}$

$V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}$

$V=3354.83m/s$

Therefore

Period T is Given as

$T=\frac{2 \pi *a}{V}$

$T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}$

$T=66262.4s$

4 0

3 years ago

Bob is threatening Tom’s life with a giant laser with wavelength (650 nm), a distance (D = 10 m) from the wall James is shackled

Fittoniya [83]

Answer:

He should stand from the center of laser pointed on the wall at 1.3 m.

Explanation:

Given that,

Wave length = 650 nm

Distance =10 m

Double slit separation d = 5 μm

We need to find the position of fringe

Using formula of distance

$d\sin\theta=n\lambda$

$d\dfrac{y}{D}=n\lambda$

$y=\dfrac{\lambda D}{d}$

Put the value into the formula

$y=\dfrac{650\times10^{-9}\times10}{5\times10^{-6}}$

$y=1.3\ m$

Hence, He should stand from the center of laser pointed on the wall at 1.3 m.

8 0

3 years ago

/Q: DHSMV definition/ i have to write 20 characters for whatever reason lol

cluponka [151]

<span><span>Department of Highway Safety and Motor Vehicles OR</span></span>

<span><span /><span><span>Division of Highway Safety and Motor Vehicles </span></span></span>

6 0

3 years ago

One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d

stira [4]

Answer:

a) Diffusion coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

$^{2} = 2Dt$ ..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

$36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr$

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

$^{2} = 2Dt$

$0.1^{2} = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds$

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

8 0

3 years ago

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.05 m/s. If the roof is pitched at

AlekseyPX

(a) The time the baseball spends in the air is 0.92 s.

(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

<h3>Time spent in air by the baseball</h3>

h = vt - ¹/₂gt²

-2.1 = (4.05 x sin 34)t - ¹/₂(9.8)(t²)

-2.1 = 2.26t - 4.9t²

4.9t² - 2.26t - 2.1 = 0

t = 0.92 s

<h3>Horizontal distance traveled by the baseball</h3>

R = Vx(t)

R = (4.05 x cos 34)(0.92)

R = 3.1 m

Thus, the time the baseball spends in the air is 0.92 s.

The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

Learn more about horizontal distance here: brainly.com/question/24784992

#SPJ1

8 0

2 years ago