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horrorfan [7]
3 years ago
11

Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double sli

ts 0.500 mm apart illuminated by 500-nm light.
Physics
1 answer:
Artist 52 [7]3 years ago
5 0

Answer:

y = 1.75 cm

Explanation:

In the double-slit experiment the equation for destructive interference is

           d sin tea = (m + ½)

λ

let's use trigonometry to find the angle

         tan θ = y / L

as all the experiment does not occur at small angles

          tan θ = sin θ / cos θ = sin θ = y / L

we substitute

        y = (m + 1/2 ) λ  L / d

we calculate

         y = (3 + ½) 500 10⁻⁹ 5.00 / 0.5 10⁻³

         y = 1.75 10⁻² m

         y = 1.75 cm

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Bella makes the 6.1m distance to her food bowl in 8.8 seconds what is her average velocity
storchak [24]
Bella’s average velocity is about 0.693 meters per second.

To find the average velocity, you must divide the distance by the change in time, which should look like v=d/t

Here is how you set up the equation-
v=6.1/8.8

Once you divide 6.1 meters by 8.8 seconds, you should get a number that looks like 0.69318182.... however, I just rounded it to 0.693 meters per second. You can round it to whatever you like.

Hope this helped! If you have any questions about what I mentioned in my answer or explanation, feel free to comment on my answer and I’ll try to get back to you!
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What is the largest ecosystem on the planet?
asambeis [7]
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What is the SI unit for graviational potential energy
scZoUnD [109]

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3 0
3 years ago
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0
3 years ago
It took 3.5 hours for a train to travel the distance between two cities at a velocity of 120 km/h. How far did the train travel
Nadusha1986 [10]

Answer: 420 km

Explanation:

120 per hour, 3.5 hours

120 x 3.5 = 420

6 0
3 years ago
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