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natka813 [3]
2 years ago
13

Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez

ers is 1000 W/m2, the same as the intensity of sunlight at the surface of the Earth. (a) What is the pressure on an atom if light from the tweezers is totally absorbed? ? Pa (b) If this pressure were exerted on a tritium atom, what would be its acceleration? (The mass of a tritium atom is 5.01 ✕ 10−27 kg. Assume the cross-sectional area of the laser beam is 6.65 ✕ 10−29 m2.)
Physics
1 answer:
Zanzabum2 years ago
6 0

(a)  3.3\cdot 10^{-6} Pa

The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by

p=\frac{I}{c}

where

I is the intensity of the wave

c is the speed of light

In this problem,

I=1000 W/m^2

and substituting c=3\cdot 10^8 m/s, we find the radiation pressure

p=\frac{1000 W/m^2}{3\cdot 10^8 m/s}=3.3\cdot 10^{-6}Pa

(b) 4.4\cdot 10^{-8} m/s^2

Since we know the cross-sectional area of the laser beam:

A=6.65\cdot 10^{-29}m^2

starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:

F=pa=(3.3\cdot 10^{-6}Pa)(6.65\cdot 10^{-29} m^2)=2.2\cdot 10^{-34}N

And then, since we know the mass of the atom

m=5.01\cdot 10^{-27}kg

we can find the acceleration, by using Newton's second law:

a=\frac{F}{m}=\frac{2.2\cdot 10^{-34} N}{5.01\cdot 10^{-27} kg}=4.4\cdot 10^{-8} m/s^2

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answer  with full explanation is attached below


3 0
2 years ago
A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust rever
Amanda [17]

Answer:

257 kN.

Explanation:

So, we are given the following data or parameters or information in the following questions;

=> "A jet transport with a landing speed

= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"

= > The distance = 425 m along the runway with constant deceleration."

=> "The total mass of the aircraft is 140 Mg with mass center at G. "

We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"

Step one: determine the acceleration;

=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.

=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.

Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).

= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).

= 257 kN.

4 0
2 years ago
Describe the distribution of wdiff in terms of its center, shape, and spread, including any plots you use
melisa1 [442]

Answer:

Figure attached

We can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.

Explanation:

We can use the following R code to obtain the data for wdiff:

source("http://www.openintro.org/stat/data/cdc.R")  #obtain the info

nrow(cdc) # number of elements

names(cdc)  # obtain the name for the variable

[1] "genhlth"  "exerany"  "hlthplan" "smoke100" "height"   "weight"   "wtdesire" "age"      

[9] "gender"  

wdiff represent the difference between desired weight (wtdesire) and current weight (weight) and we can obtain the data with the following code:

wdiff <- (cdc$weight-cdc$wtdesire)

And now we can create the histogram with this code

hist(wdiff,xlim =c(-100,150))

> mean(wdiff)

[1] 14.5891

> median(wdiff)

[1] 10

And the result is on the figure attached.

And we can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.

3 0
2 years ago
An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
aniked [119]

Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time t_{arrow} travels the distance d with a speed v_{arrow} of 40 m/s and hits the target. We can see that the equation will be:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at  340 m/s in a time t_{sound}. Obtaining :

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

t_{arrow} + t _{sound} = 1 s.

This equation allows us to write:

t _{sound} = 1 s - t_{arrow}.

Plugging this  relationship in the distance equation for the sound:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Now, we can replace d from the first equation, and obtain:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, we can just work a little bit:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Now, we can just plug this value into the first equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

6 0
2 years ago
Mrs. Brown's class is studying magnets and electricity. At the end of the unit Claire states that magnets and electricity are bo
Gwar [14]

Answer:

Magnets can create electricity and electricity can create a magnetic force.

Both electric charges and magnets do not have to touch an object in order to exert a force on it.

Electromagnets use electricity to create a magnetic force.

Explanation:

7 0
2 years ago
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