Ask question

Ask question

All categories

3 years ago

8

Suppose you watch a leaf bobbing up and down as ripples pass it by in a pond. You notice that it does two full up and down bobs

each second. What is true of the ripples on the pond?

1 answer:

otez555 [7]3 years ago

4 0

Answer:

The frequency of the ripples is 2Hz, and their period is 0.5 seconds.

Explanation:

Since the ripples on the pond are making the leaf oscillate up and down at a rate of two times per second, we can calculate the period T and the frequency f of the ripples on the pon $T=\frac{1}{0.5Hz} =$ d.

The frequency, by definition, is the number of waves per unit of time. In this case, we have two waves per second, so the frequency is 2s⁻¹, or 2Hz.

The period is the inverse of the frequency, so

$T=\frac{1}{2Hz} =0.5s$

Then, the period is equal to 0.5 seconds.

You might be interested in

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

irinina [24]

288.51 N is the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

As the system of the beam, hi_nge and cable are in equilibrium.

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The horizontal component of force + the vertical component of force

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.

Learn more about components of forces here brainly.com/question/26446720

#SPJ1

7 0

2 years ago

Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers

If a wave is traveling at a certain speed and it's frequency is doubled,what happens to the wavelength of that water

Elena L [17]

Answer:

If the frequency is doubled the wavelength is only half as long.If you put your fingertip in a pool of water and repeatedly move it up and down,you will create circular water waves.

Hope this helps!!!

4 0

2 years ago

A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.

strojnjashka [21]

Answer:

$D=99.4665307m \approx 99.5m$

Explanation:

From the question we are told that

Mass $m=6kg$

Velocity of mass $V_m=16$

Force of Tunnel $F_t=8N$

Length of Tunnel $L_t=1.6$

Height of frictional incline $H_i=2.9$

Angle of inclination $\angle =16 \textdegree$

Acceleration due to gravity $g=9.8m/s^2$

First Frictional surface has a coeﬃcient $\alpha_1 =0.21\ for\ d_c=1$

Second Frictional surface has a coeﬃcient $\alpha _2=0.1$

Generally the initial Kinetic energy is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}(6)(16)^2$

$K.E=768$

Generally the work done by the Tunnel is mathematically given as

$w_t=F_t*d_t$

$w_t=8*1.6$

$w_t=12.8J$

Therefore

$Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t$

$E_t=K.E+E_t\\E_t=768J+12.8J$

$E_t=780.8J$

Generally the energy lost while climbing is mathematically given as

$E_c=mgh$

$E_c=(6)(9.8)(2.9)$

$E_c=170.52J$

Generally the energy lost to friction is mathematically given as

$E_f=\alpha *m*g*cos\textdegree*d_c$

$E_f=0.21*6*9.8*cos16*1$

$E_f=11.86965942 \approx 12J$

Generally the energy left in the form of mass $Em$ is mathematically given as

$E_m=E_t+E_c+E_f$

$E_m=(768J)-(170.52)-(12)$

$E_m=585.48J$

Since

$E_m=\alpha_2*g*m*d$

Therefore

It slide along the second frictional region

$D=\frac{585.46}{0.1*9.81*6}$

$D=99.4665307m \approx 99.5m$

6 0

3 years ago

What change should be expected in the velocity of a body to maintain the same

Readme [11.4K]

Answer:

firstly,

ke=1÷2mv^2

on putting same ke by increasing mass by 16 times new velocity becomes v'

then

ke=1÷2×16m×v'^2

from above we can write

1÷2mv^2=1÷2×16m×v'^2

v^2=16v'^2

1÷4v=v'

hence original velocity should be decreased by 4 times to keep same ke

4 0

3 years ago