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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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According to einstein's theory of simple relativity (_E + mc(2)_). BLANK is converted into BLANK.

True [87]

Answer:

energy is converted into mass

Explanation:

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2 years ago

A proton moves at constant velocity in the direction, through a region in which there is an electric field and a magnetic field.

dimaraw [331]

Answer:

$F_{net}=0$

Explanation:

It is given that, a proton moves at constant velocity, through a region in which there is an electric field and a magnetic field such that,

The electric field is, E = 800 V/m

Magnetic field, B = 0.25 T

We know that the net force in the region of magnetic and electric field is given by Lorentz forces. But here, the proton moves with constant velocity. So, the net force acting on it is 0.

i.e.

$F_{net}=0$

Hence, this is the required solution.

7 0

2 years ago

zhannawk [14.2K]

Average speed= total distance/total time =12km/h

6 0

2 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

2 years ago

An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car

Rudik [331]

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)= $3.2 m/s^2$

Speed of car after 4.5 s

using equation of motion

v=u+at

$v=4.92+3.2\times 4.5=4.92+14.4$

v=19.32 m/s

Displacement of the car after 4.5 s

$v^2-u^2=2as$

$19.32^2-4.92^2=2\times 3.2\times s$

$349.05=2\times 3.2\times s$

s=54.54 m

4 0

2 years ago