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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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What is the anomalous expansivity of water
salantis [7]

Answer:

The anomalous expansion of water is an abnormal property of water whereby it expands instead of contracting when the temperature goes from 4°C to 0°C, and it becomes less dense. The density becomes less and less as it freezes because molecules of water normally form open crystal structures when in solid form.

7 0
3 years ago
A disk of radius R = 11 cm is pulled along a frictionless surface with a force of F = 16 N by a string wrapped around the edge.A
kenny6666 [7]

Answer: 1.76 Nm

Explanation:

If the force pulls horizontally, this means that the force is tangent to the disk at any point of the string unwinding process, so the distance d is irrelevant.

In this case, the torque is directly given by the product of the force times the distance perpendicular to the center of the disk, which is just the radius, as follows:

τ = F * r = 16 N. (0.11) m = 1.76 Nm

7 0
3 years ago
Tides occurs in oceans but not in lakes.why?​
-Dominant- [34]

Answer:

Explanation:

Tides occur in the ocean but nit in lakes because an ocean is a free flowing body of water that can travel a large area of the globe while a lake or pond only covers a small area of the earth so it is not affected by gravity as violently and in turn, prevents the formation of tides.

4 0
3 years ago
A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic
barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

(b) Change in momentum of the truck, \Delta p=m(v-u)

\Delta p=2000\ kg\times (16.11-9.44)

\Delta p=13340\ kg-m/s

Hence, this is the required solution.

6 0
3 years ago
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an
nalin [4]

Answer:

The maximum volume is 1417.87 inch^3

Explanation:

<u>Optimization Using Derivatives</u>

We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches

When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is

V=(24-2x)(30-2x)x

Operating

V=4x^3-108x^2+720x

To find the maximum value of V, we compute the first derivative and equate it to zero

V'=12x^2-216x+720=0

Simplifying by 12

x^2-18x+60=0

Completing squares

x^2-18x+81-81+60=0

(x-9)^2=21

We have two values for x

x=9+\sqrt{21}=13.58\ inch

x=9-\sqrt{21}=4.42\ inch

The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16

We'll keep only the solution

x=4.42\ inch

The width is

w=(24-2(4.42))=15.16\ inch

The length is

l=(30-2(4.42))=21.16\ inch

And the height

x=4.42\ inch

The maximum volume is

V=(15.16)(21.16)(4.42)=1417.87\ inch^3

4 0
3 years ago
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