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Leni [432]
3 years ago
7

In the lungs there are tiny sacs of air, which are called alveoli. an oxygen molecule is trapped within a sac, and the uncertain

ty in its position is 0.12 mm. what is the minimum uncertainty in the speed of this oxygen molecule?
Physics
1 answer:
AysviL [449]3 years ago
8 0

As we know that Heisenber's uncertainty law gives that

\Delta x * \Delta P = \frac{h}{4\pi}

here we know that

\Delta x = uncertainty In Position = 0.12 mm

\DeltaP = uncertainty In Momentum = m*v

now from the above formula we will have

0.12 * 10^{-3}* 32*10^{-3}v = \frac{h}{4\pi}

3.84*10^{-6}*v = \frac{6.6*10^{-34}}{4\pi}

v = 1.37 * 10^{-29}m/s

so above is the uncertainty in the speed

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Answer:

send the wagon down a higher hill

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The doppler effect is sensitive only to motion along the line of sight. <br> a. True <br> b. False
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4 0
3 years ago
A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2
andrey2020 [161]

Work needed = 23,520 J

<h3>Further explanation</h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object  

Work is the product of force with the displacement of objects.  

Can be formulated  

W = F x d  

W = Work, J, Nm  

F = Force, N  

d = distance, m  

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

5 0
3 years ago
6
Arturiano [62]

Explanation:

A light bulb changes electrical energy into <em>heat energy and light energy .</em>

4 0
3 years ago
3. Maverick and Goose are flying a training mission in their F-14. They are
Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of y_0=1500m, and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

y=y_0 - \frac{gt^2}{2}    [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

x = v.t     [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

t=\sqrt{\frac{2y_0}{g}}

Since we have y_0=1500m

t=\sqrt{\frac{2(1500)}{9.8}}

t=17.5 sec

Replacing in [2]

x = 688\ m/sec \ (17.5sec)

x = 12040\ m

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0
2 years ago
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