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2 years ago

What is the wavelength of a sound wave with a speed of 331 m/s and a frequency of 500 Hz?

Physics

1 answer:

ludmilkaskok [199]2 years ago

4 0

Answer:

0.777m

Explanation:

The sound wave has a wavelength of 0.773m.

Explanation:

To solve this problem we have to use the wave equation that is given below:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for

λ :

λ = v f

Let's plug in our given values and see what we get!

λ = 340 m s

440 s − 1

λ = 0.773 m

Hope this helps, Mark as brainliest if u want

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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de

kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

u = 0

Therefore, the equation becomes

S = 1/2 gt²

Substituting the given values in the above equation

S = 0.5 x 9.8 x 1.5²

= 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

3 0

3 years ago

If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be

il63 [147K]

Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .

If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .

It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors

1 ) charge q and

2 ) the permittivity of medium ε around .

4 0

3 years ago

a physical science textbook has a mass of 2.2 kg. If the textbook weighs 19.6 newtons on Venus, what is the strength of gravity

nadezda [96]

112244798978957866777777888844322123445545

8 0

2 years ago

Read 2 more answers

What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$