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3 years ago

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The speed of a wave changes depending on the temperature of the medium. What do you think is the reason for this?

1 answer:

igor_vitrenko [27]3 years ago

5 0

Answer: Because of the different wave speed from light and sound. Explanation: There is a major difference between the speed wave of light and sound, light travels at 186, 282 miles per second, and sound can travel at different speeds and its significantly slower so it is easier to measure it

Explanation:

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5. A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen. The 5th minimum occurs at 7.00° aw

abruzzese [7]

For a single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen,the angle is mathematically given as

theta=25.3

Option A is correct

<h3> What angle does the 18th minimum occur?</h3>

Generally, the equation for the the angle is mathematically given as

$\theta=n(\lambda/d)$

Therefore

$\theta 1/ \theta 2=n1(\lambda/d)/ n2(\lambda/d)$

In conclusion

theta/7=16/5

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2 years ago

A scientist shines light from a source onto a piece of metal, and no electrons are released by the metal. Increasing the intensi

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This is the photoelectric effect, and it is best explained by the particle model of light.

<h3>What is the photoelectric effect?</h3>

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2 years ago

A rectangle has a length of (2.5 ± 0.2) m and a width of (1.5 ± 0.2) m. Calculate the area and the perimeter of the rectangle, a

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2 years ago

Make the following conversion. 0.0097 mg = _____ g

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Your answer should be 9.7 :)

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3 years ago

Read 2 more answers

A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s

Initial velocity of bullet ( $u_{bullet}$ ) = (500 m/s)i

Speed of the bullet after collision ( $v_{bullet}$ ) = (300 m/s) j

Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

$(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}$

$(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}$

$v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j$

$v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j$

∴

The magnitude now is:

$v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}$

$\mathbf{v_{stone}= 14.6 \ m/s}$

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

$\theta = tan^{-1} (\dfrac{-7.5}{12.5})$

$\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}$

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3 years ago