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babymother [125]
3 years ago
13

Waves that move the particles of the medium parallel to the direction in which the waves are traveling are called?

Physics
1 answer:
OLEGan [10]3 years ago
8 0
<span>Longitudinal Waves is the correct answer</span>
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When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted.
olchik [2.2K]

A) We want to find the work function of the potassium. Apply this equation:

E = 1243/λ - Φ

E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function

Given values:

E = 2.93eV, λ = 240nm

Plug in and solve for Φ:

2.93 = 1243/240 - Φ

Φ = 2.25eV

B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:

E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

λ = 552nm

C) We want to find the frequency associated with the threshold wavelength. Apply this equation:

c = fλ

c = speed of light in a vacuum, f = frequency, λ = wavelength

Given values:

c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

Plug in and solve for f:

3×10⁸ = f(5.52×10⁻⁷)

f = 5.43×10¹⁴Hz

7 0
4 years ago
A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

4 0
3 years ago
A block of mass m moving due east at speed v collides with and sticks to a block of mass 2m that is moving at the same speed vv
lapo4ka [179]

Answer:

30.36°

Explanation:

By using linear momentum; linear momentum can be expressed by the relation:

mv_xi + mv_yj

where ;

m= mass

v_x = velocity of components in the x direction

v_y = velocity of components in the y direction

If we consider the east as the positive x and north as positive y which is synonymous to what we usually have on a graph.

Then;

Initial momentum = mvi + 2mvcos 45i + 2mvsin45 j

= (mv+2mvcos45)i + (2mvsin45)j

However, the masses stick together after collision and move with a common velocity: V_xi +V_yj

∴ Final momentum = 3mv (V_xi +V_yj)

= 3mV_xi + 3mV_yj

From the foregoing ;

initial momentum = final momentum

3mV_xi + 3mV_yj = (mv+2mvcos45)i+(2mvsin45j)

So;

3mV_x = mv + 2mv cos 45 \\\\3mV_y = 2mV sin 45

V_x = \frac{mv+2mvcos45 }{3m}\\\\V_x = \frac{v+2vcos45}{3}

V_y = \frac{2mvsin45}{3m} \\\\V_y = \frac{2vsin45}{3}

Finally;

The required angle θ = tan^{-1} = \frac{V_y}{V_x}

θ = tan^{-1} = \frac{\frac{2vsin45}{3}}{\frac{v+2v.cos45}{3}}

θ = tan^{-1} = \frac{2sin 45}{1+2cos45}\\\\

θ = 30.36°

7 0
3 years ago
Please help me!! I already did A,B and C but I don’t know D but if you want to help me with all then it would be great so I can
Inessa [10]

Answer:

350cm^3 =350milliters

4 0
3 years ago
Read 2 more answers
Separate sheet of paper for your answer.
evablogger [386]

Answer:

I think its B. Please keep in mind that I could be wrong.

Explanation:

4 0
3 years ago
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