All categories

4 years ago

How to find the period of a wave

1 answer:

5 0

The method to choose depends on what information you have, and
on what you can measure. Here are a few possible methods:

-- Measure the period. Start your clock when one peak
of the wave passes you. Stop the clock when the next
peak passes you. The time between the two peaks is
the wave's period.

-- Divide the wave's wavelength by its speed. That quotient
is the wave's period.

-- Use an electronic frequency meter to measure the wave's
frequency. Then take its reciprocal (divide ' 1 ' by it). The
result is the wave's period.

You might be interested in

The glowing of a neon light is caused by electrons emitting energy as they (1 point)

Lena [83]

<span>The glowing of a neon light is caused by electrons emitting energy as they </span>move from higher to lower energy levels.

5 0

4 years ago

Read 2 more answers

a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v

solniwko [45]

Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

The height of the building, h = 200 m

The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

S = ut + ½ gt²

Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)

Adding these two equations,

x + (S - x) = U tₓ

S = U tₓ

200 = 50 tₓ

∴ tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s

6 0

3 years ago

What current flows through a 2.54cm diameter rod of pure silicon that is 20cm long when 1000V is applied?

vfiekz [6]

Answer: $0.0039\ A$

Explanation:

Given

Diameter of the rod $d=2.54\ cm$

length of rod is $l=20\ cm$

Resistivity of silicon is $\rho=6.4\times 10^2\ \Omega-m$

cross-section of the rod $A$

$\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2$

Resistance of rod is R

$\Rightarrow R=\dfrac{\rho l}{A}$

$\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega$

Current is given by

$\Rightarrow I=\dfrac{V}{R}\\\\\Rightarrow I=\dfrac{1000}{25.26\times 10^4}\\\\\Rightarrow I=0.0039\ A$

3 0

3 years ago

A ball starts from rest. It rolls down a ramp and reaches the ground after 4 seconds. Its final velocity when it reaches the gro

Hunter-Best [27]

If it starts at rest the initial velocity is 0.
For an acceleration, a, and time, t, the velocity is v=at. Since at t=4, v=7, then a=7/4=1.75m/s^2

3 0

3 years ago

Read 2 more answers

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

6 0

4 years ago