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3 years ago

11

A blank star has a greater temperature than a blank star

2 answers:

zheka24 [161]3 years ago

7 0

Blue is greater than Red

DIA [1.3K]3 years ago

7 0

The answer is A) Blue, Red

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A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

<em>Now from the work-energy equivalence:</em>

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

3 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

GaryK [48]

We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

$y = \frac{m\lambda}{d}$

Here,

m = Number of order bright fringe

$\lambda$ = Wavelength

d = Distance between slits

Both distance are the same, then

$y_1 = y_2$

$\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}$

$\frac{m_1\lambda_1}{m_2\lambda_2} =1$

Rearranging to find the second wavelength

$m_1 \lambda_1 = m_2 \lambda _2$

$\lambda_2 = \frac{m_1\lambda_1}{m_2}$

$\lambda_2 = \frac{7(629)}{8}$

$\lambda_2 = 550.3nm$

Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm

7 0

3 years ago

5. A sled with no initial velocity accelerates at a rate of 5.0 m/s2 down a hill. How

ad-work [718]

Answer:

60 m/s

Explanation:

$a = \frac{vf - vi}{t}$

$5.0 = \frac{vf - 0}{12}$

$5.0 \times 12 = vf$

$60 = vf$

6 0

3 years ago

A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of $34.9^{\circ}$

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

$d sin \theta = m \lambda$

where

d is spacing between the lines in the grating

$\theta$ is the angle of the maximum

m is the order of diffraction

$\lambda$ is the wavelength of the light

For laser 1,

$d sin \theta = m_1 \lambda_1$

For laser 2,

$d sin \theta = m_2 \lambda_2$

where

$\lambda_1 = 420 nm\\\lambda_2 = 630 nm$

Since the position of the maxima in the two cases overlaps, then the term $d sin \theta$ on the left is the same for the two cases, therefore we can write:

$m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}$

Therefore:

$m_1 = 3$

$m_2 = 2$

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

$d sin \theta = m_1 \lambda_1$

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

$d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m$

$m_1 = 3$ is the order of the maximum

$\lambda_1 = 420 nm = 420\cdot 10^{-9} m$ is the wavelength of the laser light

Solving for $\theta$ , we find the angle of the maximum:

$sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572$

So the angle is

$\theta=sin^{-1}(0.572)=34.9^{\circ}$

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

5 0

4 years ago

What are some of the difficulties with measuring the volume of a gas? Select all that apply.

forsale [732]

Answer:

it is ---D because you can't measure gas and it's mass

HOPE THIS HELPS

6 0

3 years ago