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3 years ago

Please help on this one ? :)

Physics

1 answer:

Llana [10]3 years ago

3 0

If you're on the road in car 2 going 20 m/s and car 1 is going 25 m/s in the dafne direction it would look as if it was passing you at 5 m/s west. This is because 25 m/s - 20 m/s = 5 m/s

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Suppose that the speed of an electron traveling 2.0 km/s is known to an accuracy of 1 part in 105 (i.e., within 0.0010%). What i

OverLord2011 [107]

Answer: $2.89(10)^{-3} m$

Explanation:

The <u>Heisenberg uncertainty principle</u> postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.

In other words:

It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.

Mathematically this principle is written as:

$\Delta x \geq \frac{h}{4 \pi m \Delta V}$ (1)

Where:

$\Delta x$ is the uncertainty in the position of the electron

$h=6.626(10)^{-34}J.s$ is the Planck constant

$m=9.11(10)^{-31}kg$ is the mass of the electron

$\Delta V$ is the uncertainty in the velocity of the electron.

If we know the accuracy of the velocity is $0.001\%$ of the velocity of the electron $V=2 km/s=2000 m/s$ , then $\Delta V$ is:

$\Delta V=2000 m/s(0.001\%)$

$\Delta V=2000 m/s(\frac{0.001}{100})$

$\Delta V=2(10)^{-2} m/s$ (2)

Now, the least possible uncertainty in position $\Delta x_{min}$ is:

$\Delta x_{min}=\frac{h}{4 \pi m \Delta V}$ (3)

$\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)}$ (4)

Finally:

$\Delta x_{min}=2.89(10)^{-3} m$

5 0

3 years ago

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

3 0

3 years ago

What type of relationship does this graph show?

aliya0001 [1]

Answer:

I'm pretty sure it shows a direct relationship

Explanation:

5 0

3 years ago

A certain electric circuit obeys Ohm's law. If the resistance of the circuit is doubled, what will happen to the current through

netineya [11]

Answer:

currrent will be halved

Explanation:

v = ir

v/r = i multiply both sides by 1/ 2

v / (2r) = 1/2 i

4 0

3 years ago

A 65 Kg roller-blade is accelerating at 5 m/s/s across the side walk. What force would be necessary for this acceleration to occ

Neporo4naja [7]

Newton's second law states that the force applied to an object is equal to the product between the mass m of the object and its acceleration a:
$F=ma$
Using $m=65 kg$ and $a=5 m/s^2$ , we can find the value of the force applied to the roller-blade to obtain this acceleration:
$F=(65 kg)(5 m/s^2)=325 N$

3 0

3 years ago