Is Their A Video On The Pysical Thingie

What is the frequency of a wave that has a wavelength of 0.39 m and a speed

gogolik [260]

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

$f = \frac{c}{ \lambda} \\$

where

c is the velocity

From the question

wavelength = 0.39 m

c = 86 m/s

We have

$f = \frac{86}{0.39} \\ = 220.512820...$

We have the final answer as

Hope this helps you

7 0

3 years ago

One of the harmonics of a string fixed at both ends has a frequency of 52.2 Hz and the next higher harmonic has a frequency of 6

Angelina_Jolie [31]

Solution :

Frequency may be defined as the number of observation or number of waves that is taken in per unit time. The unit of frequency is Hertz or Hz.

It is given that :

Successive harmonic frequencies, f = 52.2 Hz

and f' = 60.9 Hz

Therefore, fundamental frequency, F = f' - f

F = 60.9 - 52.2

F = 8.7 Hz

Therefore the string which is fixed at both the ends forms all the harmonics.

6 0

2 years ago

the area of 1 end of U-tube is 0.0 metre square and that of the other end is 1 metre square when you force was applied on the li

Lapatulllka [165]

Answer:

The question is wrong Since if you apply Force on 0.0m²It would mean That the pressure exerted=F/A=F/0

An since we can't divide a number by 0, the question is wrong

6 0

3 years ago

A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =

stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N , Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

$d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }$

$d_{13} =\sqrt{24.68}$ * 10⁻²m = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) = - 3.67*10⁻⁵ N

F₂₃x = F₂₃ = +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x = - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

$F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2} }$

$F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2} }$ *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction (α)

$\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )$

$\alpha =tan^{-1}( \frac{-3.67 }{5.71} )$

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0

3 years ago

If sensed fuel in the Piper Archer reaches ____ gallons in either tank, the yellow Low Fuel caution light will illuminate on the

lana66690 [7]

Answer:

10 litres.

Explanation:

The air fuel ratio for aircraft is 12:1. It is due to simplicity piper archer is equipped with continuous flow injection system. There is a fuel sensing is Piper Archer which informs the pilot about the low fuel when the fuel level reaches to as low as 10 litres only in the tank. The yellow low furl light will illuminate to alert the pilot.

3 0

3 years ago