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lisabon 2012 [21]
3 years ago
12

The uncertainty in the position of an electron along an x axis is given as 53 pm, which is about equal to the radius of a hydrog

en atom. What is the least uncertainty in any simultaneous measurement of the momentum component px of this electron?
_____ kg·m/s
Physics
1 answer:
Jobisdone [24]3 years ago
4 0

To solve this problem it is necessary to apply the concepts related to the Heisenberg's uncertainty principle. Under this principle we understand the relationship that the minimum range of error in position (x) times the minimum range of error in momentum (p) is, at a minimum, about equal to the Planck constant, mathematically that is,

\Delta p = \frac{h}{\Delta x}

Replacing with our values we have,

\Delta p = \frac{6.63*10^{-34}J\cdot s}{2\pi (53*10^{-12}m)}

\Delta p = 1.99*10^{-24} kg\cdot m/s

Therefore the least uncertainty in any simultaneous measurement of the momentum component px of this electron is 1.99*10^{-24} kg\cdot m/s

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the pull of gravity on mars is 3.7m/s^2. if a astronaut on mars lifts a 10 kg rock 1 m off the ground, just to see whats under i
Elanso [62]
Gravitational potential energy can be calculated using the formula:

PE_{grav} =mgh

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know. 
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it. 
PE_{grav} =mgh
PE_{grav} =(10kg)(3.7m/s^{2})(1m)
PE_{grav} =37J


7 0
4 years ago
2. Jak myślisz, dlaczego w naszym słowniku bez problemu odnajdziemy następujące wyrażenia
Juliette [100K]

Answer:

..........,

Explanation:

6 0
4 years ago
Which of these is the best explaination for why two negatively charged balloons move apart without ever touching?
Dennis_Churaev [7]
Hello, sorry this is a little late!

I believe the correct answer to your question would be option D, <span>electric charges have electric fields surrounding them to allow them to exert forces on other objects without touching them.

I just took this test, and can 100% confirm this is the proper answer.

Hope this helps, and have a great day! :)</span>
6 0
3 years ago
Read 2 more answers
Tres litros de oxigeno gaseoso a 15 grados centígrados y a presión atmosférica (1atm), se lleva a una presión de 10mm de Hg. ¿ c
lana [24]

Answer:

Three liters of gaseous oxygen at 15 degrees Celsius and at atmospheric pressure (1atm), is brought to a pressure of 10mm Hg. What will be the volume of the gas now if the temperature has not changed?

Explanation:

Given that, the temperature is constant

Then, using gay Lussac law

P1•V1 / T1 = P2•V2 / T2

Since temperature is constant

Then, T1 = T2 = T and they cancels out

So, we are left with

P1•V1 = P2•V2

Given that, .

Initial volume

V1 = 3 litres.

Initial pressure

P1 = 1atm = 101325 Pa

Final pressure

P2 = 10mmHg = 1333.22 Pa

Then, we want to find the final volume V2

Make V2 subject of formula.

V2 = V1•P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litres

So, the final volume is 288 litres.

In Spanish

Dado que la temperatura es constante

Luego, usando la ley gay de Lussac

P1 • V1 / T1 = P2 • V2 / T2

Como la temperatura es constante

Entonces, T1 = T2 = T y se cancelan

Entonces, nos quedamos con

P1 • V1 = P2 • V2

Dado que, .

Volumen inicial

V1 = 3 litros.

Presión inicial

P1 = 1atm = 101325 Pa

Presión final

P2 = 10 mmHg = 1333.22 Pa

Entonces, queremos encontrar el volumen final V2

Hacer V2 sujeto de fórmula.

V2 = V1 • P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litros

Entonces, el volumen final es de 288 litros

8 0
4 years ago
There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think
dangina [55]

Answer:

4.44 rpm

Explanation:

\omega = Angular speed

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Europa = \frac{3138000}{2}\ m

R = Radius of arm = 6 m

The acceleration due to gravity is given by

g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

a_c=\omega^2R

a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s

Converting to rpm

1\ rad/s=\frac{60}{2\pi}\ rpm

0.46547\ rad/s=0.46547\times \frac{60}{2\pi}\ rpm=4.44\ rpm

The angular speed of the arm is 4.44 rpm

8 0
3 years ago
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