All categories

3 years ago

What are 3 wave properties?

2 answers:

8 0

Amplitude: the height of the wave
wavelenght: the distance betweeen adjacent crests
period: the time it takes for one complete wave to pass a given point

Dmitrij [34]3 years ago

4 0

The three wave properties include:
Amplitude, height of the wave measured in meters.

Wavelength, the distance measured between crests, measured in meters.

Period, the time it takes for the wave to finish it's cycle, measured in seconds.

Hope this is the answer you were looking for and that it helps!! :)

You might be interested in

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago

What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?

joja [24]

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

= ${\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2}$ = 3984378 m / $s^{2}$

Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

= ${\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2}$ = 3984378 m / $s^{2}$

3 0

3 years ago

If a car has a mass of 200 kilograms and produces a force of 500 N, how fast will the car accelerate?

Hitman42 [59]

Answer:

2.5m/s²

Explanation:

Given parameters:

Mass of car = 200kg

Force on car = 500N

Unknown:

Acceleration of the car = ?

Solution:

According to Newton's second law of motion:

Force = mass x acceleration

Insert the given parameters and find the acceleration;

500 = 200 x acceleration

acceleration = 2.5m/s²

3 0

3 years ago

Read 2 more answers

What is the best reason to talk to an adult if you are feeling symptoms of a mental disorder? They can help you find services in

vitfil [10]

<span>They can help you find services in your community.</span>

5 0

3 years ago

Read 2 more answers

You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe agai

QveST [7]

Answer:

Explanation:

Given mass of grindstone $m=90\ kg$

radius of stone $r=0.34\ m$

angular speed of disc $\omega =90\ rpm$

Steel Axle applying a force of $F=20\ N$

coefficient of kinetic friction $\mu =0.2$

Frictional Torque applied by steel is given by

$\tau=r\times f_r$

$\tau =r\times \mu F$

where $f_r$ =frictional force

$\tau =r\times \mu \times F$

$\tau =0.34\times 0.2\times 20$

$\tau =1.36\ N-m$

Torque is also given by

$\tau =I\cdot \alpha$

where $\alpha$ =angular acceleration

I=moment of Inertia

$\tau =0.5Mr^2\times \alpha$

$0.5Mr^2\times \alpha =1.36\ N-m$

$0.5\times 90\times 0.34^2\times \alpha =1.36$

$\alpha =0.261\ rad/s^2$

3 0

3 years ago