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2 years ago

What type of stars will potentially one day

Physics

1 answer:

xz_007 [3.2K]2 years ago

4 0

Answer:

2, High mass stars.

Add-on:

i hope this helped at all. im sorry if its incorrect.

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A parked car's horn (belonging to a musician) emits a concert A of frequency 440 on a day when the speed of sound is 342 m/s. Yo

julsineya [31]

Answer:

19.08 m/s

Explanation:

f = actual frequency emitted by the parked car's horn = 440 Hz

V = speed of sound = 342 m/s

f' = frequency of the horn observed by you = 466 Hz

v = speed of your car moving towards the parked car = ?

frequency of the horn observed by you is given as

$f' = \frac{Vf}{V - v}$

$466 = \frac{(342)(440)}{342 - v}$

v = 19.08 m/s

3 0

3 years ago

A car accelerates from rest at 3.00 m/second squared What is the velocity after 5 secs ? What is the displacement after 5 second

andrey2020 [161]

Explanation:

after 5 seconds, the velocity is (5s)(3m/s²) = 15m/s

The displacement after 5s is

x=vo + (1/2)at²

x = 0 + (1/2)(3m/s²)(5s)(5s)

x= 37.5 m

3 0

3 years ago

The tuning fork has a frequency of 426.7 is found to cause resonance in a closed column of air measuring 0.186 (the diameter of

lidiya [134]

Answer:

The velocity of the tuning fork sound is 317.4648 m/s

Explanation:

The given parameters are;

The frequency of the fork, f = 426.7 Hz

The length of the closed air column, L = 0.186 m

The diameter of the tube, d = 0.15 m

At the fundamental frequency in a closed tube, we have;

λ = 4·L

Where;

λ = The wavelength of the wave

L = The length of the tube

From the equation for the velocity of a wave, we have;

v = f·λ

Where;

v = The velocity of the (sound) wave

f = The frequency of the wave = 426.7 Hz

λ = 4·L = 4 × 0.186 m = 0.744 m

∴ v = 426.7 Hz × 0.744 m = 317.4648 m/s

Therefore, the velocity of the sound produced by the tuning fork, v = 317.4648 m/s

3 0

3 years ago

The pupil of a cat's eye narrows to a slit width of 0.5 mm in daylight. What is the angular resolution of the cat's eye in dayli

Elenna [48]

Answer:

C. 10⁻³ rads

Explanation:

Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:

θ = λ/a

where,

θ = angular resolution of Cat's eye = ?

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m

Therefore,

θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)

Therefore,

θ = 0.001

θ = Sin⁻¹(0.001)

θ = 0.001 rad = 1 x 10⁻³ rad

Hence, the correct answer is:

4 0

3 years ago

The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specif

Neko [114]

Hey again!

Ok..

Now... The melting Point of this solid is 90°C.

Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

You can say its a Solid Liquid Mixture.

Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.

Q= Q' + Q"

Q= mc∆0 + ml

∆0 = 90°c - 30°c = 60°c

Q= 2.5(390)(60) + (2.5)(4000)

Q=6.9 x 10⁴Joules

7 0

3 years ago