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3 years ago

You drive 773 miles east, then turn around and drive 773 miles west. What is your displacement, to the nearest mile?

Physics

1 answer:

Oliga [24]3 years ago

8 0

Answer:0miles

Explanation:

Since they are moving opposite direction we say

Displacement =773-773

Displacement =0miles

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Given the initial wavefunction Ψ (x, 0) = Axexp (-k x) withx> 0 andk> 0, and Ψ (x, 0) = 0 forx <0, what value must A ta

madam [21]

Answer with explanation:

The Normalization Principle states that

$\int_{-\infty }^{+\infty }f(x)dx=1$

Given

$f(x)=xe^{-kx}(x>0\\\\0(x$

Thus solving the integral we get

$\int_{0 }^{+\infty }A\cdot xe^{-kx}dx=1\\\\A\int_{0 }^{+\infty }\cdot xe^{-kx}dx=1$

The integral shall be solved using chain rule initially and finally we shall apply the limits as shown below

$I=\int xe^{-kx}dx\\\\x\int e^{-kx}dx-\int \frac{d(x)}{dx}\int e^{-kx}dx\\\\-\frac{xe^{-kx}}{k}-\int 1\cdot \frac{-e^{-kx}}{k}\\\\\therefore I=\frac{e^{-kx}}{k}-\frac{xe^{-kx}}{k}$

Applying the limits and solving for A we get

$I=\frac{1}{k}[\frac{1}{e^{kx}}-\frac{x}{e^{kx}}]_{0}^{+\infty }\\\\I=-\frac{1}{k}\\\\\therefore A=-k$

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3 years ago

A ball is thrown straight up with a velocity of 16 m/s; what will be its velocity 2.0s after being released?

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The answer for the question is 8m/s

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3 years ago

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Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

Explanation:

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3 years ago

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3 years ago

Read 2 more answers

A satellite orbits the Earth in the same direction it rotates in a circular orbit above the equator a distance of 250 km from th

UkoKoshka [18]

Answer:

Clock on the satellite is slower than the one present on the earth = 29.376 s

Given:

Distance of satellite from the surface, d = 250 km

Explanation:

Here, the satellite orbits the earth in circular motion, thus the necessary centripetal force is provided by the gravitation force and is given by:

$\frac{mv^{2}}{R} = \frac{GMm}{R^{2}}$

where

v = velocity of the satellite

R = radius of the earth = 6350 km = 6350000 m

G = gravitational constant = $6.674\times 10^{- 11} m^{3}/ks-s^{2}$

M = mass of earth = $5.972\times 10^{24} kg$

Therefore, the above eqn can be written as:

$v = \sqrt{\frac{GM}{R}}$

Now, for relativistic effects:

$\frac{v}{c} = \sqrt{\frac{GM}{Rc^{2}}} = 26.41\times 10^{- 6}$

Now,

r = R + 250

$\frac{v_{surface}}{c} = {\frac{1}{c}\frac{2\pi R}{24} = 1.54\times 10^{-6}$

Ratio of rate of satellite clock to surface clock:

$\frac{\sqrt{1 - \frac{v^{2}}{c^{2}}}}{\sqrt{1 - \frac{v_{surface}^{2}}{c^{2}}}} = 3.43\times 10^{-10}$

Clock on the satellite is slower than the one present on the earth:

$3.43\times 10^{-10}\times 24\times 3600 = 29.376 s$

4 0

3 years ago