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3 years ago

A wave has a frequency of 450 Hz and a wavelength of 0.52 m. What is the

Physics

1 answer:

Vikki [24]3 years ago

6 0

Answer:

A. 230 m/s

Explanation:

The speed of a wave is given by:

$v=f \lambda$

where

f is the frequency

$\lambda$ is the wavelength

For the wave in the problem,

f = 450 Hz (frequency)

$\lambda=0.52 m$ (wavelength)

Substituting into the equation, we find the speed:

$v=(450)(0.52)=234 m/s \sim 230 m/s$

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HELP ASAP TIMED TEST

balu736 [363]

Answer:

Correct choice: b 4H

Explanation:

Conservation of the mechanical energy

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

$U_1 = mgH$

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

$\displaystyle U_2=\frac{1}{2}mv^2$

Since the energy is conserved, U1=U2

$\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]$

For the speed to be double, we need to drop the snowball from a height H', and:

$\displaystyle mgH'=\frac{1}{2}m(2v)^2$

Operating:

$\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]$

Dividing [2] by [1]

$\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}$

Simplifying:

$\displaystyle \frac{H'}{H}=4$

Thus:

H' = 4H

Correct choice: b 4H

4 0

3 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: