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Aneli [31]
3 years ago
12

Which equations represent the relationship between wavelength and frequency for a sounds wave

Physics
1 answer:
ioda3 years ago
6 0

   

Waves can be described using a number of different characteristics of a wave. Wavelength and frequency are two such characteristics. The relationship between wavelength and frequency is that the frequency of a wave multiplied by its wavelength gives the speed of the wave

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What is measurement?​
Semmy [17]

Answer:

How long or wide something is

Explanation:

5 0
3 years ago
The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able
vovangra [49]

Answer:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

q_e=-1.6\cdot 10^{-19}\ C

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is 5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C

Let's test the possible charges listed in the question:

-8.0 \times 10 ^{-19 }. We have just found it's a possible charge of a particle

-3.2 \times 10 ^{-19 }. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

-1.2 \times 10 ^{-19 } this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 } cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge -4.8 \times 10 ^{-19 }\ C is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

5 0
3 years ago
A jumbo jet weighing 65,000 kg Is accelerating at 3 m/s^2 while its jet engines are putting out 200000 N of thrust.
sveticcg [70]

Answer:

Fa = 5000 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

Let's assume that the movement of the plane is to the right, any movement or force to the right will be marked with a positive sign, while any force or movement to the left, will be taken as negative.

The force of the turbine drives the plane to the right, therefore it is positive, the acceleration is constant and keeps the movement to the right, therefore it is positive, the wind drag force tries to prevent the movement of the plane to the left therefore it is negative, with this analysis we deduce the following equation.

ΣF = m*a

where:

ΣF = sum of forces [N] (units of Newtons)

m = mass = 65000 [kg]

a = acceleration = 3 [m/s²]

Fa = force exerted by the air [N]

200000 - Fa = 65000*3

Fa = 200000 - (3*65000)

Fa = 5000 [N]

8 0
3 years ago
What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa
dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}

E=3.315\times10^{-17}\ J

(b).  We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}

E=6.709\times10^{-21}\ J

(c). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}

E=3.315\times10^{-11}\ J

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}

E=6.709\times10^{-9}\ J

Hence, This is the required solution.

6 0
3 years ago
You can find electric power lines under the ground by looking for magnetic fields at ground level. This is best explained by whi
luda_lava [24]
What was the answer?

8 0
3 years ago
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