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3 years ago

8

What happens to the velocity of a sound wave in air if the temperature of the air increases? A. velocity increases B. velocity d

oes not change C. velocity decreases D. velocity drops to zero

2 answers:

Drupady [299]3 years ago

5 0

Answer:

Velocity increases for plato users

Explanation:

tigry1 [53]3 years ago

3 0

The answer of your question is A.

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In a double slit experiment, 450 nm light passes through two slits producing an interference pattern where the first bright frin

musickatia [10]

Answer:

1.3 x 10⁻⁴ m

Explanation:

$\lambda$ = wavelength of the light = 450 nm = 450 x 10⁻⁹ m

n = order of the bright fringe = 1

θ = angle = 0.2°

d = separation between the slits

For bright fringe, Using the equation

d Sinθ = n $\lambda$

Inserting the values

d Sin0.2° = (1) (450 x 10⁻⁹)

d (0.003491) = (450 x 10⁻⁹)

d = 1.3 x 10⁻⁴ m

6 0

2 years ago

Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

or

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

or

$f=\frac{33.84}{0.935}$

or

$f=36.19\approx 36\ \textup{Hz}$

5 0

2 years ago

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

2 years ago

Someone please help its a simple power problem.

SOVA2 [1]

Well 200 doubled or (x2)=400 if that’s what it means

7 0

2 years ago

A lead nucleus is spherical with a radius of about 7 ✕ 10⁻¹⁵ m. The nucleus contains 82 protons (and typically 126 neutrons). Be

Komok [63]

Answer:

∈= $3.1584x10^{26} \frac{V}{m}$

Explanation:

Using the Gauss Law to determine the electric field of the net flux at the surface of the nucleus

∈ $=\frac{P}{E_{o}}$

The P is the charge density and 'Eo' is the constant of permittivity in free space

to find P

$P=\frac{q}{V}$

$V=\frac{4}{3}*\pi*r^3$

$V=\frac{4}{3}\pi*(7x10^{-15})^3$

$V=2.932x10^{-14} m^3$

$P=\frac{82C}{2.932x10^{-14}m^3}=2.7965x10^{15} \frac{C}{m^3}$

So replacing

∈ $=\frac{2.7965x10^{15}\frac{C}{m^3}}{8.8542x10^{-12}\frac{C^2}{N*m^2}}$

∈= $3.1584x10^{26} \frac{V}{m}$

3 0

3 years ago