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inna [77]
3 years ago
10

The uncertainty in the position of an electron along an x axis is given as 5 x 10-12 m. What is the least uncertainty in any sim

ultaneous measurement of the momentum component px of this electron?
Physics
1 answer:
Vsevolod [243]3 years ago
4 0

Answer:

The least uncertainty in the momentum component px is 1 × 10⁻²³ kg.m.s⁻¹.

Explanation:

According to Heisenberg's uncertainty principle, the uncertainty in the position of an electron (σx) and the uncertainty in its linear momentum (σpx) are complementary variables and are related through the following expression.

σx . σpx ≥ h/4π

where,

h is the Planck´s constant

If σx = 5 × 10⁻¹²m,

5 × 10⁻¹²m . σpx ≥ 6.63 × 10⁻³⁴ kg.m².s⁻¹/4π

σpx ≥ 1 × 10⁻²³ kg.m.s⁻¹

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A molecule of roughly spherical shape has a mass of 6.10 × 10-25 kg and a diameter of 0.70 nm. The uncertainty in the measured p
Ilia_Sergeevich [38]

Answer:

\Delta v = 0.123 m/s

Explanation:

As per the law of uncertainty we know that

\Delta p \times \Delta x = \frac{h}{4\pi}

now we know that

\Delta x = 0.70 nm

m = 6.10 \times 10^{-25}

also we have

h = 6.626 \times 10^{-34} J.s

now we will have

m\Delta v \times \Delta x = \frac{h}{4\pi}

(6.10 \times 10^{-25})\Delta v \times (0.70 \times 10^{-9}) = \frac{6.626 \times 10^{-34}}{4\pi}

\Delta v= 0.123 m/s

4 0
3 years ago
Coins were developed as a medium of exchange because other items like cows, grain, and land were more difficult to move from pla
miv72 [106K]
True, they used them because its easier to trade coins than products
7 0
3 years ago
Read 2 more answers
30
pickupchik [31]

Answer:

22m/s

Explanation:

lowest part on the graph (closest to x-axis)

4 0
2 years ago
the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity
NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
3 years ago
Assuming a typical efficiency for energy use by the body, how many slices of pizza must you eat to walk for 2.5 h at a speed of
larisa86 [58]

Answer:

2.7 Pizzas.

Explanation:

The power required to walk through 5km in 1 hour is 380W.

A watt is basically Jules per second, then we need to standardized this measurement to second.

5km/hr is equal to,

\frac{5km}{hr}*\frac{1hr}{3600s}*\frac{1000m}{1km}=1.389m/s

Walking by 2.5 hours is equal to a distance of,

d=v*t=1.389*(2.5*3600) = 12500m

The total energy required then would be,

E = \frac{380J}{1.389m/s}(12500)=3.4199*10^6J

Then we know that one pizza slice gives 1260*10^3J of energy, the total pizza needed are,

\eta = \frac{3.4199*10^6}{1260*10^3} = 2.7142

<em>Then you need to buy 3 pizza.</em>

6 0
3 years ago
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