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3 years ago

Two similar waves X and Y travel through the same medium. Wave X has a frequency of 200 hertz and a

Physics

1 answer:

velikii [3]3 years ago

4 0

Answer:

The correct solution is option A "10 meters".

Explanation:

The given values are:

Wave X,

Frequency,

$F_x=200 \ H_z$

Wavelength

$\lambda_x=35 \ m$

Wave Y,

Frequency,

$F_y=700 \ Hz$

Let its wavelength = $\lambda_y$

As we know,

⇒ $v=f \lambda$

For both waves, medium is same then

⇒ $v_x=v_y$

⇒ $f_x \lambda_x=f_y \lambda_y$

then,

⇒ $\lambda_y=\frac{f_x \lambda_x}{f_y}$

On substituting the estimated values in the above equation, we get

⇒ $=\frac{200\times 35}{700}$

⇒ $=\frac{7000}{700}$

⇒ $=10 \ m$

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3 years ago

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Sometimes students are taught that "air expands as it is heated". And sometimes they are taught that "as air expands it cools".

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Answer:

Both are true under specific circumstances. And are related to Boyle's law. volume and pressure in a gas are inversely proportional.

Explanation:

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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

An airplane flies with a velocity of 750. kilometers

viktelen [127]

Answer:

650 km/hr

Explanation:

Draw a right triangle from (0.0) (Point A) down 30 degrees and to the right for a length of 750 (Point B). Then draw a line from B up to the x axis to make a right angle (Point C). Use the cosine function to find line AC, the vector portion of AB that lies of the x (East) axis. Cosine(30)= Adjacent/Hypotenuse.

Cos(30) = AC/750

750*(cos(30)) = AC

AC = 649.5 km/hr

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