All categories

3 years ago

If frequency is kept constant, how are the velocity and wavelength of a wave related?

2 answers:

4 0

Wave speed = (frequency) x (wavelength)

If the frequency is kept constant, then the wavelength
is directly proportional to the wave speed.

Elan Coil [88]3 years ago

3 0

<h3>Answer;</h3>

Velocity and wavelength are directly proportional when frequency is kept constant.

<h3>Explanation;</h3>

Frequency of a wave is the number of complete oscillations made by a given wave in one second.
Wavelength on the other hand, is the distance between two successful crests or troughs in a transverse wave or two successful rarefactions or compressions in a longitudinal waves.
The speed of a wave is given by the product of the frequency of a wave and the wavelength.
Speed = Frequency × wavelength, 
Therefore, if frequency is kept constant, then the speed of a wave is directly proportional to the wavelength, such that an increase in wavelength increases the speed of the wave and vice versa.

You might be interested in

A car, starting from the origin, travels

sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

8 0

3 years ago

Read 2 more answers

A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They

soldier1979 [14.2K]

Answer:

a. $\thta=71^{\circ}$

b. $v_f=3.78 m/s$

Explanation:

We are given that

$m_1=53 kg$

$v_1=2.9 m/s$

$m_2= 72 kg$

$v_2=6.2 m/s$

a.We have to find the angle

$\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}$

$\theta=71^{\circ}$

b. We have to find the speed $v_f$

According to law of conservation of momentum

$m_1v_1=(m_1+m_2)v_fcos\theta$

$53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f$

$v_f=\frac{153.7}{40.7}=3.78 m/s$

3 0

3 years ago

A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel

Salsk061 [2.6K]

Answer:

If the particle is an electron $E_y = 3.311 * 10^3 N/C$

If the particle is a proton, $E_y = 6.08 * 10^6 N/C$

Explanation:

Initial speed at the origin, $u = 3 * 10^6 m/s$

$\theta = 38^0$ to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, $m_e = 9.1 * 10^{-31} kg$

Mass of a proton, $m_p = 1.67 * 10^{-27} kg$

The electric field intensity along the positive y axis $E_y$ , can be given by the formula:

$E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\$

If the particle is an electron:

$E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C$

If the particle is a proton:

$E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 6.08 * 10^6 N/C$

8 0

3 years ago

In a certain region of space, the electric potential is V(x,y,z)=Axy-Bx^2+Cy , where A,B , and C are positive constants.a) Calcu

laiz [17]

Answer:

a) Eₓ = - A y + 2B x , b) Ey = -Ax –C , c) Ez = 0 , d) The correct answer is 3

Explanation:

The electric field and the electric power are related

E = - dV / ds

a) Let's find the electric field on the x axis

Eₓ = - dV / dx

dV / dx = A y - B 2x

Eₓ = - A y + 2B x

b) calculate the electric field on the y-axis

Ey = - dV / dy

dV / dy = A x + C

Ey = -Ax –C

c) the electric field on the z axis

dv / dz = 0

Ez = 0

.d) at which point the electric field is zero

Since the electric field is a vector quantity all components must be zero

X axis

0 = = - A y + 2B x

y = 2B / A x

Axis y

0 = -Ax –C

.x = -C / A

We substitute this value in the previous equation

.y = 2B / A (-C / A)

.y = 2 B C / A2

The correct answer is 3

6 0

4 years ago

You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 35 m/s. The ball has __________________

Elden [556K]

Answer:

The ball has kinetic energy

the kinetic energy is 945 J

Explanation:

4 0

3 years ago

Read 2 more answers