All categories

3 years ago

Calculate the frequency of the wave shown below.

Physics

1 answer:

jenyasd209 [6]3 years ago

5 0

$\color{skyblue}{ \underline{ \frak { \: option \: ( \: c \: ) = 2 \: hertz ✓}}}$

$\: \:$

Given :

Wavelength ( λ ) = 2 m

$\: \:$

Speed = 4 m/s

$\: \:$

We, have to find frequency :

$\:$

$\large \tt \: Frequency = \frac{Speed}{Wavelength ( \: λ \: )}$

$\: \:$

$\large \tt \: Frequency = \frac{4}{2}$

$\: \:$

$\large \tt \: Frequency = \cancel \frac{4}{2}$

$\: \:$

$\pink{ \boxed{\large \tt \: Frequency =2 \: Hertz ✓}}$

$\: \:$

Hope Helps!

You might be interested in

at Niagara Falls, if 505 kg of water fall a distance of 50.0 m, what is the increase in the internal energy of the water at the

anygoal [31]

Answer:

Hope that helps

Explanation:

the increase in the internal energy of the water at the bottom of the falls

= mgh

= 505 * 9.81 * 50

= 247703 J ----answer

5 0

3 years ago

Please answer as fast as possible

sineoko [7]

Explanation:

the pressure is given by P=hpg

where h is height,p is density, g is 10m/s²

thus P= 0.05m•13.6x10³•10

P=6.8x10³ Hgmm

3 0

3 years ago

Two one-coulomb point charges are at a distance of 1.67 m from each other. What is the size of the electric force between the tw

pychu [463]

Explanation:

The given data is as follows.

distance (r) = 1.67 m

$q_{1} = q_{2}$ = 1 C

It is known that formula for electric force between two charges is as follows.

F = $\frac{kq_{1} \times q_{2}}{r^{2}}$

Hence, putting the given values into the above formula as follows.

$\frac{9 \times 10^{9} \times 1 \times 1}{(1.67)^{2}}$

= $3.227 \times 10^{9}$

As it is given that mass of Falcon Heavy rocket is 1421. Therefore, weight of 1 Falcon rocket is as follows.

weight of 1 falcon rocket = $142 \times 10^{3} \times 9.8$

Therefore, number of falcon rockets will be calculated as follows.

No. of falcon rockets = $\frac{3.227 \times 10^{9}}{142000 \times 9.8}$

= $2.31897 \times 10^{3}$

= 2318.9 falcons

Thus, we can conclude that force between the two point charges above is equal to the weight of 2318.9 Falcon Heavy rockets.

5 0

3 years ago

Which volcanic hazard can block the sunlight and temporarily cool the Earth’s surface?

BaLLatris [955]

If a volcano epulses massive amounts of dust into the atmosphere, those two things will/can happen.
The events will last until the dust lays down on the earth.

8 0

3 years ago

Read 2 more answers

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$

δm is the error of the mass value.
δd is the error of the diameter value.
δh is the error of the length value.

Let's find each partial derivative:

1. $\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089$

2. $\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004$

3. $\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071$

Therefore:

$\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}$

$\delta\rho=0.00058$

So the density is:

$\rho=(0.0089\pm 0.00058) kg/cm^{3}$

I hope it helps you!

3 0

3 years ago