All categories

2 years ago

The blackbody curve for a star named Zeta is shown below. What is the peak wavelength for this star?

Physics

1 answer:

g100num [7]2 years ago

4 0

We are asked for the x coordinate of vertex

Here vertex is present somehow at 850-870nm

The most applicible answer should be 860nm

Option C is correct

You might be interested in

What is a plasma?A.A superheated liquidB.A solid in which the particles vibrate very quicklyC.A gas in which the atoms have star

Mandarinka [93]

plasma is a superheated liquid

So, a would-be the correct option.

5 0

1 year ago

What is the most significant challenge for organisms that live in estuaries?

marysya [2.9K]

"Changing water salinity" is the most significant challenge for organisms that live in estuaries.

<u>Answer:</u> Option D

<u>Explanation:</u>

For estuaries, alkalinity levels are usually the maximum at a river's mouth where the ocean water falls for, and the minimum upstream where freshwater falls in. Although salinity vary throughout the tidal cycle. In estuaries, salinity rates usually decrease in spring as snow melt and rain raises the freshwater flow from streams and groundwater.

It influences the chemical environments within the estuary, especially the dissolved oxygen (DO) levels in the water. The level of oxygen that would get dissolved in water or its solubility get declined when the alkalinity rises.

8 0

4 years ago

For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total fo

Snezhnost [94]

Answer:

114.86%

Explanation:

In both cases, there is a vertical force equal to the sprinter's weight:

Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r)² + (mg)²

F² = m² ((v²/r)² + g²)

F = m √((v²/r)² + g²)

Compared to just the vertical force:

F / Fy

m √((v²/r)² + g²) / mg

√((v²/r)² + g²) / g

Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√((12²/26)² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).

5 0

3 years ago

Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having

igomit [66]

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

$x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s$

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) $\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s$

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

6 0

3 years ago

Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of

pantera1 [17]

Explanation:

Given that,

Linear speed of both disks is 5 m/s

Mass of disk 1 is 10 kg

Radius of disk 1 is 35 cm or 0.35 m

Mass of disk 2 is 3 kg

Radius of disk 2 is 7 cm or 0.07 m

(a) The angular velocity of disk 1 is :

$v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s$

(b) The angular velocity of disk 2 is :

$v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s$

(c) The moment of inertia for the two disk system is given by :

$I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2$

Hence, this is the required solution.

6 0

3 years ago