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3 years ago

(Hurry please) A student drops a rubber ball onto a surface. Assume that

Physics

2 answers:

tensa zangetsu [6.8K]3 years ago

8 0

Answer:

B. The ball makes a sound with each bounce, suggesting that some

kinetic energy transformed into sound.

Explanation:

horsena [70]3 years ago

6 0

Answer:

I think it's Answer A. Because the ball eventually came to a stop, infering that the energy was used up. Hope this helped! :)

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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:

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:

$m_1 = 3$

$m_2 = 2$

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

It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i

timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is $4.8 \cdot 10^{-6}\,mg$

Explanation:

The ration between Uranium mass and total sample mass is: $\frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}$

For a sample of mass 1.2 mg, the amount of uranium is:

$1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg$

7 0

3 years ago

Two 6 ohm resistors in parallel gives an equivalent resistance of

Mashcka [7]

Answer:

3 ohms

Explanation:

6×6/6+6 =3 .............

3 0

3 years ago

What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s

AlexFokin [52]

Answer:

$K=0.023J$

Explanation:

From the question we are told that:

Mass $m=0.15$

Velocity $v=0.5m/s$

Angular Velocity $\omega=8.4rad/s$

Generally the equation for Kinetic Energy is mathematically given by

$K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)$

$K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))$

$K=0.023J$

8 0

3 years ago

This question is about the alpha particle experiment

Wewaii [24]

This suggest the the atom has a very small positively charged nucleus in the mass of the atom is concentrated.
And the electrons revolves around the nucleus in their orbits.

4 0

4 years ago