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finlep [7]
3 years ago
8

Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is foc

used to a very small-diameter spot. Because small particles are attracted to regions of high light intensity, the focused beam can be used to "grab" onto particles and manipulate them for various experiments. In one experiment, a 10 mW laser beam is focused to a spot that has a diameter of 0.67 μm
Part A: What is the intensity of the light in this spot?

Part B: What is the amplitude of the electric field?
Physics
1 answer:
vekshin13 years ago
8 0

Answer:

Explanation:

Part A) Using

light intensity I= P/A

A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B)  Using

I=c*ε*E^2/2

rearrange to solve for E= \sqrt{((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

I= the already solved light intensity= 8.85*10^10 W/m^2

amplitude of the electric field E= \sqrt{(9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= \sqrt{(1.8*10^11) / (2.66*10^-3) = \sqrt{(6.8*10^13) = 8.25*10^6 V/m    

 

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Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an
krok68 [10]

Answer:

A) 31 kJ

B)  1.92 KJ

C) 40 ,  2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈  0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.013) )

          = 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh  hence  F = mgh / d

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.345) )

          = 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

=  31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

  Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48

4 0
3 years ago
Someone please help? :)
Gnoma [55]

Answer:

(A) F_N - mg\cos\theta

Explanation:

The net force perpendicular to the surface of the incline is the sum of the gravity force component, which is mgcos(theta), and the reactionary normal force caused by the surface of the incline. The sum is F_N - mgcos(theta) and is usually 0 which is why the object is not moving perpendicularly to the surface of the incline.

8 0
4 years ago
True or False: Heavy elements such as Carbon and Nitrogen, that are necessary for life were created in the Big
Orlov [11]

Answer:

i believe the answer is true

Explanation:

everything was created in the big bang

8 0
3 years ago
If a hiker that weighs 600 newtons climbs a 50 meter hill, how much gravitational potential energy has the hiker gained?
Illusion [34]

The gravitational potential energy U is defined as the product of mass m, the acceleration of gravity g and the height of object h.

U = mgh

We do not have the mass of the hiker. But we know that its W weight is:

W = mg

Where

g = 9.8 m/s^2

So:

m = \frac{W}{g}\\\\m = \frac{600}{g}.

So:

U = (\frac{600}{g})gh\\\\U = (600N)(50m)

U = 30000J

The hiker has gained 30,000 J of energy

6 0
3 years ago
A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top
Anton [14]

Answer:

y_{max} = 0.829\,m

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}

v = 2.913\,\frac{m}{s}

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}

After some algebraic handling, a second-order polynomial is formed:

12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s

1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0

The roots of the polynomial are, respectively:

\Delta s_{1} \approx 0.128\,m

\Delta s_{2} \approx -0.099\,m

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

\Delta s \approx 0.128\,m

The maximum height that the block reaches after rebound is:

(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}

y_{max} = 0.829\,m

4 0
3 years ago
Read 2 more answers
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