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S_A_V [24]
3 years ago
5

I need help with this

Physics
1 answer:
Vilka [71]3 years ago
3 0

The right answer is atom. Please rate brainiest.

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A sound wave enters a new medium where sound travels faster. How does this affect the frequency and wavelength of the sound?
iren [92.7K]
The frequency doesn't change. If the wavespeed increases, then the wavelength must also increase ... It's just the distance the wave travels during each complete wiggle.
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Which of the following describes the variable that responds to a change?
timama [110]
Dependent variable is your answer.
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A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light,"
Mashutka [201]

The solution for this problem is:

If they feel 50% of their weight that means that the centripetal force is also 50% of their weight 1g - 0.5g = 0.5g 


Then 0.5* 9.8m/s² * 18m = 88.2 would be v² 

Then get the square root, the answer would be:
and v = 9.391 m/s is the answer.

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3 years ago
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An interference pattern is produced by light with a wavelength 580 nm from a distant source incident on two identical parallel s
irakobra [83]

Answer:

Explanation:

1 )

Here

wave length used that is λ = 580 nm

=580 x 10⁻⁹

distance between slit d = .46 mm

= .46 x 10⁻³

Angular position of first order interference maxima

= λ / d radian

= 580 x 10⁻⁹ / .46 x 10⁻³

= 0.126 x 10⁻² radian

2 )

Angular position of second order interference maxima

2 x  0.126 x 10⁻² radian

= 0.252 x 10⁻² radian

3 )

For intensity distribution the formula is

I = I₀ cos²δ/2 ( δ is phase difference of two lights.

For angular position of θ1

δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

= I₀ X .998

For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

=  I₀ cos².126x 10⁻²

8 0
3 years ago
One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p
Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}

kg\dfrac{m^2}{s^2}=Joule

P=\dfrac{kgm^2}{s^3}

So, the answer is D. the proper replacement unit for one joule per second

8 0
3 years ago
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