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4 years ago

What is found by multiplying frequency by wavelength?

2 answers:

5 0

That's the speed of the wave.

("Velocity" is not a good way to say it, because 'velocity' always needs a direction,
and neither frequency nor wavelength tells anything about direction.)

Viktor [21]4 years ago

3 0

Velocity= frequency x wavelength

You might be interested in

Two point charges, q1 = 2.0 × 10−7 C and q2 = −6.0 × 10−8 C, are held 25.0 cm apart. (a) What is the electric field at a point 5

AlladinOne [14]

Answer:

a) $432000\frac{N}{C}\hat{i}$

b) $-6.92\times10^{-14}N\hat{i}$

Explanation:

The magnitude of the electric field generated by a charged particle at a distance r is:

$E= k\frac{|Q|}{r^{2}}$

With Q the charge of the particle and k the constant ()

So, the electric field generated by q1 knowing that the point 5.0 cm apart the negative charge is $25.0cm-5.0cm=20.0 cm=0.2m$ apart the positive charge is:

$E_1= k\frac{|q1|}{r_1^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|2.0\times10^{-7}|}{(0.2)^{2}}$

$E_1= 45000\frac{N}{C}$

and the electric field generated by q2:

$E_2= k\frac{|q2|}{r_2^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|-6.0\times10^{-8}|}{(0.05)^{2}}$

$E_2=216000\frac{N}{C}$

Those are the magnitudes of the electric field, but electric field is a vector quantity, so the direction is important. Electric field generated by negative particles points towards the charge and electric field generated by positive particles points away the particle. So, if we define positive direction towards negative particle (x-axis):

$\overrightarrow{E_2}=+216000\frac{N}{C}\hat{i}$

$\overrightarrow{E_1}= +45000\frac{N}{C}\hat{i}$

Vector quantities satisfy superposition principle, this is $\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$ , with E the total electric field.

$\overrightarrow{E}=(216000+45000)\hat{i}=432000\frac{N}{C}\hat{i}$

b) The force is:

$\overrightarrow{F}=e*\overrightarrow{E}$ ,

with q the charge of an electron

$\overrightarrow{F}=(-1.61\times10^{-19})*(432000)\hat{i}=-6.92\times10^{-14}N\hat{i}$

8 0

3 years ago

Helpppppppp ill give brainliest

Nina [5.8K]

Answer:

I believe the answer is

$B. \: 3 \: red, \: 8 \: yellow, \: 1 \: blue$

Explanation:

I may be incorrect because I used to do this a long time ago but I believe I am correct

HOPE THIS HELPS!

4 0

3 years ago

Attached here is the question:

ICE Princess25 [194]

Answer:

t = 6 [s]

Explanation:

In order to solve this problem we must first use this equation of kinematics.

$v_{f}^{2}=v_{o}^{2} +2*a*x$

where:

Vf = final velocity = 0 (the car comes to rest)

Vo = initial velocity = 72 [km/h]

a = acceleration [m/s²]

x = distance = 60 [m]

First we must convert the velocity from kilometers per hour to meters per second.

$72 [\frac{km}{h}]*\frac{1000m}{1km} *\frac{1h}{3600s} =20 [m/s]$

$0=(20)^{2} -2*a*60\\400 = 120*a\\a=3.33[m/s^{2} ]$

Now using this other equation of kinematics.

$v_{f}=v_{o}-a*t$

0 = 20-3.33*t

t = 6[s]

5 0

3 years ago

If I push on the wall with 75 Newtons of force, the wall will push back with ______

Sauron [17]

Answer:

(F)reaction = - 75 N

where, negative sign shows opposite direction.

Explanation:

This question can be answered using Newton's third law of motion. The Newton's Third Law of Motion states that for every action force there is an equal but opposite reaction force.

(F)action = - (F)reaction

Hence, in our scenario if we consider the 75 Newton force applied on the wall to be the action force then the reaction force of the wall must be equal to it in opposite direction. Therefore, the reaction push of the wall must be equal to 75 N.

(F)reaction = - 75 N

where, negative sign shows opposite direction.

6 0

3 years ago

A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel

dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0

4 years ago