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3 years ago

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In a single-slit diffraction experiment, the slit is 0.1mm wide and is illuminatedby monochromatic light with a wavelength of 50

0 nm. The diffraction patternis observed on a screen 10 m away. What is the spacing between successiveminima in the pattern.

1 answer:

dem82 [27]3 years ago

4 0

Answer:

$0.1$ m

Explanation:

$d$ = slit width = 0.1 mm = 0.1 x 10⁻³ m

$\lambda$ = wavelength of monochromatic light = 500 nm = 500 x 10⁻⁹ m

$D$ = Distance of the screen = 10 m

$w$ = Spacing between successive minima

Spacing between successive minima is given as

$w = \frac{2D\lambda }{d}$

Inserting the values given

$w = \frac{2(10)(500\times10^{-9}) }{0.1\times10^{-3}}$

$w = 0.1$ m

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A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal

ss7ja [257]

Answer:

$v_{o}=141.51m/s$

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

$x=1420m\\g=-9.8m/s^{2}$

From the equation of x-position we know that

$x=v_{ox}t=v_{o}cos(35)t$

Solving for $v_{o}$

$v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)}$ (1)

Now, if we analyze the equation of y-position we got

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

At the end of the motion y=0

$0=v_{o}sin(35)t+\frac{1}{2}gt^{2}$

Knowing the equation for $v_{o}$ in (1)

$0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}$

$\frac{1}{2}(9.8)t^{2}=1420tan(35)$

Solving for t

$t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s$

Now, we can solve (1)

$v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s$

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Liz rushes down onto a subway platform to find her train already departing. she stops and watches the cars go by. each car is 8.

Snezhnost [94]

The average velocity can be calculated using the formula:

v = d / t

For the 1st car, the velocity is calculated as:

v1 = 8.60 m / 1.80 s = 4.78 m / s

While that of the 2nd car is:

v2 = 8.60 m / 1.66 s = 5.18 m / s

Now we can solve for the acceleration using the formula:

v2^2 = v1^2 + 2 a d

Rewriting in terms of a:

a = (v2^2 – v1^2) / 2 d

a = (5.18^2 – 4.78^2) / (2 * 8.6)

a = 0.23 m/s

Therefore the train has a constant acceleration of about 0.23 meters per second.

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4 years ago

Consider a small child sliding down two differently shaped slides that have the same coefficient of kinetic friction. How must t

timama [110]

Answer:

Having a bigger angle above the horizontal

Explanation:

Applying the energy conservation theorem:

$K_1+U_1+W_{ext}=K_2+U_2$

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The friction force is given by:

$F_f=F_N*\µ$

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So when the angle of the slide is bigger, the friction force decreases, for example:

for 45 degrees:

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for 75 degrees:

$F_N=m.g*cos(75)\\F_N=0.26(m.g)\\$

as you can see if the angle is bigger above the horizontal, the friction force is reduced and so the work done by that force. We didn't have to change the height of the slide, so the potential gravitational energy remains the same.

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