All categories

2 years ago

What is the uncertainty of the position of the bacterium? express your answer with the appropriate units?

1 answer:

4 0

For two un-related quantities, the Heisenberg uncertainty equations holds: the prduct of the two uncertainty quantities is greater than $\hbar/2$
Example of unrelated quantities are position and momentum, energy and time.
Thus
$\Delta x*\Delta p \ \textgreater \ \hbar/2$
Knowing the speed of the bacteria the uncertainty in its position is
$\Delta x \ \textgreater \ \hbar/(2 \Delta p) =\hbar/(2mv)$

You might be interested in

When a brick is dropped from the roof of a house, it lands on the ground with a speed v. What will the speed at ground level be

Nina [5.8K]

Twice as much by the first brick

4 0

3 years ago

Which property increases as an electromagnetic waves energy decreases

Rufina [12.5K]

Answer:

Wavelength

Explanation:

3 0

2 years ago

Read 2 more answers

What effect does temperature have on the composition of ocean water?

Burka [1]

The answer is letter B. XD

4 0

3 years ago

Read 2 more answers

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

andrey2020 [161]

Considering that the book is moving with constant speed, the force applied by Anna must be the same that the friction force:

$F = F_R = F\cdot \mu_k\cdot N$

If we clear the previous equation:

$\mu_k = \frac{F}{F_R} = \frac{19.5\ N}{51.7\ N} = \bf 0.38$

5 0

2 years ago

Read 2 more answers

g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th

kati45 [8]

The ball has height y and velocity v at time t according to

y = v₀ t - 1/2 g t ²

and

v = v₀ - g t

where v₀ is its initial speed and g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time t such that

-2.72 m = v₀ t - 1/2 g t ²

-7.94 m/s = v₀ - g t

Solve for t in the second equation:

t = (v₀ + 7.94 m/s)/g

Substitute this into the first equation and solve for v₀ :

-2.72 m = v₀ (v₀ + 7.94 m/s) /g - 1/2 g ((v₀ + 7.94 m/s)/g)²

-2.72 m = v₀²/g + (7.94 m/s) v₀/g - 1/2 (v₀ + 7.94 m/s)²/g

2 (-2.72 m) g = 2v₀² + 2 (7.94 m/s) v₀ - (v₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2v₀² + (15.9 m/s) v₀ - (v₀² + (15.9 m/s) v₀ + 63.0 m²/s²)

-53.3 m²/s² = v₀² - 63.0 m²/s²

v₀² = 9.73 m²/s²

v₀ = 3.12 m/s

3 0

2 years ago