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3 years ago

A giant star has a luminosity of 300 its color is most likely to be

1 answer:

amm18123 years ago

5 0

<span>The correct answer is blue. If you look at a luminosity star chart, called the Hertzsprung Russell Diagram, you will see the measurement of luminosity on the left side, and you will see a curve of stars with different colors (which correlate to the colors of the stars). Look for 30 on the luminosity measurement (look between 1 and 100). Then, move horizontally across the diagram until you hit the stars, whose color will be blue. Thus, blue is the answer.</span>

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In which scenario is an animal doing work? Check all that apply.

AleksAgata [21]

Answer:

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6 0

2 years ago

A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle

Yuri [45]

Answer:

(a). Index of refraction are $n_{red}$ = 1.344 & $n_{violet}$ = 1.406

(b). The velocity of red light in the glass $v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

The velocity of violet light in the glass $v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

Explanation:

We know that

Law of reflection is

$n_1 \sin\theta_{1} = n_2 \sin\theta_{2}$

Here

$\theta_1$ = angle of incidence

$\theta_2$ = angle of refraction

(a). For red light

1 × $\sin 56.6$ = $n_{red}$ × $\sin 38.4$

$n_{red}$ = 1.344

For violet light

1 × $\sin 56.6$ = $n_{violet}$ × $\sin 36.4$

$n_{violet}$ = 1.406

(b). Index of refraction is given by

$n = \frac{c}{v}$

$n_{red}$ = 1.344

$v_{red} = \frac{c}{n_{red} }$

$v_{red} = \frac{3(10^{8} )}{1.344}$

$v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

$v_{violet} = \frac{3(10^{8} )}{1.406}$

$v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

This is the velocity of violet light in the glass.

8 0

3 years ago

In which point the electric intensity of a sphere is maximum

g100num [7]

Here, as the charge is uniformly distributed in the sphere, we will consider s as an area of the sphere which is, s=4πr2 and r is radius of the gaussian surface shown in the figure above. From this, it can be seen that

7 0

3 years ago

The pressure in Arabian sea is much bigger than near the surface?<br> please don't scam

sashaice [31]

Answer:

In general, the annual sea surface temperatures(SSTs) in the Bay of Bengal(BOB) are higher than the Arabian sea(AS). because, there are two main reasons for higher SST in the Bay of Bengal than the Arabian Sea. they are 1. stratification, 2.strong mixing

stratification is nothing but a phenomenon which stratifies(layers) the sea water when different density water(fresh water, rain water) add into the sea water. So the stratification in the bay of Bengal is comparatively high than the Arabian sea due to the high river discharge and precipitation in the BOB than the AS. the mixing process over the Arabian sea is higher than the Bay of Bengal due to the prevailing of strong winds over the AS (because of the presence of the mountains of east Africa) than Bay of Bengal (because of the winds over the BOB are sluggish in nature then the AS). But generally winds over the sea mixes easily the normal sea water than stratified/stabilized sea water column. That's why less mixing will takes place over the surface of BOB than the AS. So due to the presence of less mixing over the surface of the Bay of Bengal than the Arabian sea, the SST values over the Arabian sea are always lower than the Bay of Bengal. that's why the Arabian sea is colder than the Bay of Bengal.

Explanation:

4 0

3 years ago

Read 2 more answers

What force is necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius 0.7 m with a period of 0.5 s? What i

inn [45]

Answer:

88.34 N directed towards the center of the circle

Explanation:

Applying,

F = mv²/r................... Equation 1

F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.

But,

v = 2πr/t................... Equation 2

Where t = time, π = pie

Substitute equation 2 into equation 1

F = m(2πr/t)²/r

F = 4π²r²m/t²r

F = 4π²rm/t²............. Equation 3

From the question,

Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s

Constant: π = 3.14

Substitute these values into equation 3

F = 4(3.14²)(0.7)(0.8)/0.5²

F = 88.34 N directed towards the center of the circle

8 0

3 years ago