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3 years ago

what will happen when two wave pulses that are the same size approach one another from opposite directions?

1 answer:

5 0

When two or more waves meet, they interact with each other. The interaction of waves with other waves is called wave interference. Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude.

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A 65 kg cart travels at a constant speed of 4.6 m/s. What is its kinetic energy?

Talja [164]

Answer:

Explanation:

mass (m) = 65 kg

velocity (v) = 4.6 m/s

Kinetic energy (KE)

= 1/2 * m * v²

= 1/2 * 65 * 4.6²

= 687.7 J

hope it helps :)

7 0

3 years ago

An image of a car with height of 14 cm occurred in the mirror which is located at a T-

Thepotemich [5.8K]

Answer:

The distance of car form the mirror is 330 cm.

Explanation:

height of object, h = 140 cm

height of image, h' = 14 cm

radius of curvature, R = 60 cm

focal length, f = R/2 = + 30 cm

Let the distance of image is v and the distance of object is u.

$\frac{h'}{h}\frac{v}{u}\\\\\frac{14}{140} =\frac{v}{u}\\\\v =\frac{u}{10}$

Use the formula of focal length

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{30}=\frac{10}{u}+\frac{1}{u}\\\\\frac{1}{30}=\frac{11}{u}\\\\u = 330 cm$

3 0

3 years ago

John see Hua running towards him at 11m/s. While running, Hua throws a ball to john at 5m/s. What is the speed of the ball as ob

Sav [38]

6 m/s HUUUUUUUUUUUAAAAAAAAAAAAAAAAAAAAAAAH

7 0

3 years ago

A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture f/4.00 to photograph a bear t

labwork [276]

Answer:

<h3>(A) The width $x = 0.24 \times 10^{-3}$ m</h3><h3>(B) The new width is $1.32 \times 10^{-3}$ m</h3>

Explanation:

Given :

Focal length $f = 135 \times 10^{-3} m$

Maximum aperture $D = \frac{f}{4}$

Wavelength $\lambda = 550 \times 10^{-9}$ m

(A)

From rayleigh criterion,

$\theta = \frac{1.22 \lambda }{D}$

$\theta =\frac{ 1.22 \times 550 \times 10^{-9} }{33.75 \times 10^{-3} }$

$\theta = 1.98 \times 10^{-5}$ rad

From angle formula,

$x = R\theta$

Where $R =$ 12 m ( given in example )

$x = 12 \times 1.98 \times 10^{-5}$ m

$x = 23.76 \times 10^{-5}$

$x = 0.24 \times 10^{-3}$ m

(B)

We know that $\theta$ is proportional to the $x$ and inversely proportional to the $D$

so we write the new width, here $x$ is 5.5 times larger than above case

$x = 0.24 \times 10^{-3} \times \frac{22}{4}$

$x = 1.32 \times 10^{-3}$ m

6 0

3 years ago

After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress strain curve and calcul

Fittoniya [83]

Answer:

Part a: <em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

Part b: <em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

Part c: <em>The value of Young's modulus at given point is 172 GPa.</em>

Part d: <em>The percentage elongation is 18.55%.</em>

Part e: The percentage reduction in area is 15.81%

Explanation:

From the complete question the data is provided for various Loads in ductile testing machine for a sample of d0=20 mm and l0=40mm. The plot is drawn between stress and strain whose values are calculated using following formulae. The corresponding values are attached with the solution.

The engineering-stress is given as

$\sigma=\frac{F}{A}\\\sigma=\frac{F}{\pi \frac{d_0^2}{4}}\\\sigma=\frac{F}{\pi \frac{(20 \times 10^-3)^2}{4}}\\\sigma=\frac{F}{3.14 \times 10^{-4}}$

Here F are different values of the load

Now Strain is given as

$\epsilon=\frac{l-l_0}{l_0}\\\epsilon=\frac{\Delta l}{40}\\$

So the curve is plotted and is attached.

<em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

<em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

Young's Modulus is given as

$E=\frac{\sigma}{\epsilon}\\E=\frac{238 /times 10^6}{0.00138}\\E=172,000 MPa\\E=172 GPa$

<em>The value of Young's modulus at given point is 172 GPa.</em>

The percentage elongation is given as

$Elongation=\frac{l_f-l_0}{l_0} \times 100\\Elongation=\frac{47.42-40}{40}\times 100\\Elongation=18.55 \%\\$

So the percentage elongation is 18.55%

The reduction in area is given as

$Reduction=\frac{A_0-A_n}{A_0} \times 100\\Reduction=\frac{\pi \frac{d_0^2}{4}-\pi \frac{d_n^2}{4}}{\pi \frac{d_0^2}{4}}\times 100\\Reduction=\frac{{d_0^2}-{d_n^2}}{{d_0^2}} \times 100\\Reduction=\frac{{20^2}-{18.35^2}}{{20^2}} \times 100\\Reduction=15.81\%$

So the reduction in area is 15.81%

6 0

3 years ago