A sleeping 68 kg man has a metabolic power of 79 w .

g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb

Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g

Buoyant force on lead = .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115 / 33.5

= .36

= 36 %

4 0

2 years ago

Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to p

Bad White [126]

Answer:

Diode Lasers

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250

microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is

4. The optical gain bandwidth (as measured between half intensity points) will

normally depend on the pumping current (diode current) but for this problem

assume that it is 2 nm.

(a) What is the mode integer m of the peak radiation?

(b) What is the separation between the modes of the cavity? Please express your

answer as Δλ.

(c) How many modes are within the gain band of the laser?

(d) What is the reflection coefficient and reflectance at the ends of the optical

cavity (faces of the InGaAsP crystal)?

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the

beam is a Gaussian beam with the waist located at the output. And the beam

waist size is approximately the x-y dimensions of the cavity.)

Solution:

(a) The wavelength λ of a cavity mode and length L are related by

n

mL

2

λ = , where m is the mode number, and n is the refractive index.

So the mode integer of the peak radiation is

1290

1055.1

10250422

6

= ×

××× == −

−

λ

nL

m .

(b) The mode spacing is given by nL

c f 2

=Δ . As

λ

c f = , λ

λ

Δ−=Δ 2

c f .

Therefore, we have nm

nL f

c

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .

(c) Since the optical gain bandwidth is 2nm and the mode spacing is 1.2nm, the

bandwidth could fit in two possible modes.

For mode integer of 1290, nm

m

nL 39.1550

1290

10250422 6

= ××× ==

−

λ

Take m = 1291, nm

m

nL 18.1549

1291

10250422 6

= ××× ==

−

λ

Or take m = 1289, nm

m

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

8 0

2 years ago

What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle θ fro

netineya [11]

Refer to the diagram shown below.

v = the tangential speed.
r = the radius of the horizontal circle.
T = tension in the string.
θ = the angle that the string makes with the vertical
m = Bob's mass (mg = the weight)
F = centripetal force
l = the length of the string

From geometry,
r = l sin θ

The centripetal acceleration is
$a= \frac{v^{2}}{r} = \frac{v^{2}}{l \,sin \theta}$
The centripetal force is
$F = \frac{m v^{2}}{l \, sin \theta}$

For vertical force balance,
T cosθ = mg (1)
For horizontal force balance,
$Tsin \theta = F = \frac{m v^{2}}{l sin \theta}$ (2)
Divide (2) by (1).
$tan \theta = \frac{v^{2}}{gl sin \theta} \\\\ v^{2} = gl\, sin \theta \,tan \theta \\\\ v= \sqrt{gl \, sin \theta \, tan \theta}$

Answer: $v = \sqrt{gl \, sin \theta \, tan \theta}$

5 0

2 years ago

Read 2 more answers

The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

VladimirAG [237]

We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force is mathematically given as

F=\frac{F}{dx}

Therefore

F=-kdx

k=600Nm^{-1}

now

K.E=0.5x ds^2

K.E=600*(-0.1^2)

K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

For more information on this visit

brainly.com/question/23379286

6 0

1 year ago

A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to draw b

lapo4ka [179]

The best conclusion to draw based on the description would be: <span>A.The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
This phenomenon happened because </span><span>The electric field from a positive charge will points away from the charge while the electric field from a negative charge will points toward the charge</span>

4 0

2 years ago

Read 2 more answers