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Vladimir79 [104]

3 years ago

11

1. A student practicing for a track meet ran 263 m in 30 sec. What was her average speed?

2 answers:

serg [7]3 years ago

8 0

The correct answer is: 8.3 m/s

Subtract. then, divide by 2.

Debora [2.8K]3 years ago

5 0

1) 8.3 m/s :)))))))))))))))

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A car starts from rest and accelerates at 5 m/s/s.

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Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago

What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?

Vlada [557]

Answer:

$Ep=-2.3*10^{-19}J$

Explanation:

The change in potential energy can be expressed as:

$Ep=K.\frac{q1.q2}{r}$

where K is a constant with a value of $9*10^{9}\frac{N.m^{2}}{C^{2}}$ , q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is $+1.6*10^{-19}C$ and the charge for the electron is $-1.6*10^{-19}C$ .

Converting r=1.0nm to m:

$1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m$

Replacing values:

$Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}$

$Ep=-2.3*10^{-19}J$

5 0

3 years ago