All categories

2 years ago

During destructive interference, two waves moving through the same medium will

Physics

2 answers:

ddd [48]2 years ago

6 0

They will amplify eachother.

Phoenix [80]2 years ago

5 0

It will create a smaller amplitude

You might be interested in

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.

svetlana [45]

Explanation:

The given data is as follows.

Electric field between plates without dielectric, $E_{1} = 3.50 \times 10^{5} V/m$

Electric field between the plates with dielectric, $E_{2} = 2.40 \times 10^{5} V/m$ .

Permittivity of free space, $\epsilon_{o}$ = $8.85 \times 10^{-12} C^{2}/Nm^{2}$

Now, we will determine the charge density as follows.

$\sigma_{i} = \epsilon_{o}(E_{1} - E_{2})$

= $8.85 \times 10^{-12} \times (3.50 \times 10^{5} - 2.40 \times 10^{5})$

= $9.735 \times 10^{-7} C/m^{2}$

Thus, we can conclude that the charge density on each surface of the dielectric is $9.735 \times 10^{-7} C/m^{2}$ .

6 0

2 years ago

A bicycle has a momentum of 50.00 kg·m/s and a velocity of 5.0 m/s. What is the bicycle’s mass?

-BARSIC- [3]

Answer:

10 kilograms

Explanation:

The formula for mass

mass (kg)= momentum ( p) divided by velocity (v)

therefore this translates to 50 divided by 5

which gives the final answer as <u>10kgs</u>

7 0

2 years ago

With mechanical waves, what is moving and what stays in roughly the same place?

Zielflug [23.3K]

The part that moves are called anti-nodes. The stationary pars are nodes

6 0

3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

A 50 N girl climbs the flight of stairs in 3 seconds. How much work does she

kvv77 [185]

She uses 0 power because she simply has no gas left. She used all of it getting a 50 pc McDonald’s nugget with a side of ranch and a medium fry and a large water.

7 0

3 years ago