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3 years ago

8

A wave emitted from a source has a frequency of 10 Hz and wavelength 2.5 m. How much time will it take to reach a person located

5 m from the source?

1 answer:

const2013 [10]3 years ago

4 0

Answer:

time taken by the wave to reach the person is 0.2 s

Explanation:

As we know that the speed of the wave is given as

$v = \lambda f$

here we know that the wavelength of the wave is

$\lambda = 2.5 m$

$f = 10 Hz$

now speed of the wave is given as

$v = 10(2.5)$

$v = 25 m/s$

Now time taken by the wave to reach 5 m distance is

$t = \frac{L}{v}$

$t = \frac{5}{25}$

$t = 0.2 s$

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Car A with a mass of 725 kilograms is traveling east at an initial velocity of 15 meters/second. It collides head–on with car B,

ikadub [295]

Answer:

$p_t_o_t_a_l=250kg\frac{m}{s}$

Explanation:

<u>The total momentum of a system is defined by:</u>

$(mv)_t_o_t=m_1v_1+m_2v_2+...$

Where,

$(mv)_t_o_t$ is the total momentum or it could be expressed also as $p_t_o_t_a_l$ .

$m_1$ and $m_2$ represents the masses of the objects interacting in the system.

$v_1$ and $v_2$ are the velocities of the objects of the system.

<em>Remember: </em><em>The momentum is a fundamental physical magnitude of vector type.</em>

We have:

$m_1=725 kg$

$v_1=15\frac{m}{s}\\m_2=625 kg$

We are going to take the east side as positive, and the west side as negative. Then the velocity of the car B, has to be <u>negative</u>. It goes in a different direction from car A.

$v_2=-17\frac{m}{s}$

Then the total momentum of the system is:

$p_t_o_t_a_l=m_1v_1+m_2v_2\\p_t_o_t_a_l=(725kg)(15\frac{m}{s})+(625kg)(-17\frac{m}{s})\\p_t_o_t_a_l=10875kg\frac{m}{s}-10625kg\frac{m}{s}\\p_t_o_t_a_l=250kg\frac{m}{s}$

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3 years ago

The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of

kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

$\rho_{0\°C}= 730 kg/m^{3}$ is the density of gasoline at $0\°C$

$\beta=9.60(10)^{-4} \°C^{-1}$ is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to $m^{3}$ :

$V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}$ (1)

Knowing density is given by: $\rho=\frac{m}{V}$ , we can find the mass $m_{1}$ of 8.50 gallons:

$m_{1}=\rho_{0\°C}V$

$m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg$ (2)

Now, we have to calculate the factor $f$ by which the volume of gasoline is increased with the temperature, which is given by:

$f=(1+\beta(T_{f}-T_{o}))$ (3)

Where $T_{o}=0\°C$ is the initial temperature and $T_{f}=21.7\°C$ is the final temperature.

$f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))$ (4)

$f=1.020832$ (5)

With this, we can calculate the density of gasoline at $21.7\°C$ :

$\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)$

$\rho_{21.7\°C}=745.207 kg/m^{3}$ (6)

Now we can calculate the mass of gasoline at this temperature:

$m_{2}=\rho_{21.7\°C}V$ (7)

$m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})$ (8)

$m_{2}=24.070 kg$ (9)

And finally calculate the mass difference $\Delta m$ :

$\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg$ (10)

$\Delta m=0.4911 kg$ (11) This is the extra mass of gasoline

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3 years ago

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VLD [36.1K]

Answer:

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1 year ago

Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl

Trava [24]

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

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Wab = q ( Va - Vb )

Wac = q ( Va - Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

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= q [ ( Va-Vc ) - ( Va - Vb )]

= Wac - Wab

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2 years ago

Which of the following statements describes a compound machine?

ASHA 777 [7]

From the word compound, the compound machine is already a combination of two or more types of simple machine. Thus, the answer is letter C. Because of its complexity, it is able to perform several other functions than a simple one.

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2 years ago

Read 2 more answers