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3 years ago

The 10-kg uniform rod is pinned at end

1 answer:

3 0

Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left
Sum moments about the pivot to zero.
10.0(9.81)[(2sinθ)/2] + 50 - 60(2sinθ)[2cosθ] = 0 98.1sinθ + 50 - (120)2sinθcosθ = 0 98.1sinθ + 50 - (120)sin(2θ) = 0
by iterative answer we discover that
θ ≈ 0.465 radians
θ ≈ 26.6º

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When a car driving up a hill with constant speed: I. its kinetic energy is decreasing. II. its potential energy is constant. III

madam [21]

A car driving up a hill at a constant speed experiences no change in its kinetic energy while it's potential energy increases with increasing height, thus none of the options are correct.

Understanding the concept

Consider a car moving up the hill at a constant speed as shown in the figure below. The following forces act on the car:

N is the normal reaction force acting in an upward direction

f_s is the static friction force exerted due to friction between the road and the tires of the car
f_k is the rolling friction force in the direction opposing that of the tire
mg is the force acting in a downward direction.

θ is the angle of inclination.

Here as the car is moving up the hill at a constant speed, the net force exerted on the car is zero. Also, the kinetic energy of the car will not change as its velocity is constant and the potential energy will change with increasing height. Thus, none of the given options are correct.

Learn more about motion on an incline here:

<u>brainly.com/question/13513083</u>

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5 0

2 years ago

Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the s

AURORKA [14]

Answer:

Phase difference = pi/4 radians

Explanation:

Given:

- The wavelength of incident light λ = 600 nm

- The split separation d = 0.85 mm

- Distance of screen from split plane L = 2.8 m

Find:

What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?

Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

sin ( Q ) = m*λ /d

m = d*sin(Q) / λ

- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

r = sqrt(2.8^ + 0.0025^) = 2.8

sin(Q) = 0.0025 / 2.8

Hence. m = 0.00085*0.0025 / 2.8*(600*10^-9)

m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer: Phase difference = pi/4 radians

6 0

3 years ago

An airplane is .68 Kilometers long. How many Millimeters long is the plane?

xxTIMURxx [149]

Answer:

680000

mark brainliest

4 0

2 years ago

Read 2 more answers

Hurryyyyy pleaseeeee!!!!!

kolbaska11 [484]

Answer: 2.2x10^4

Explanation: in a adiabatic process pV^y = constant.

So V2=3V1 and p2 = p1 / 3^1.67 = 2.2x10^4 Pa

4 0

3 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.

(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

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

= ×

××× == −

−

m .

(b) The mode spacing is given by nL

c f 2

=Δ . As

c f = , λ

Δ−=Δ 2

c f .

Therefore, we have nm

nL f

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .

bandwidth could fit in two possible modes.

For mode integer of 1290, nm

nL 39.1550

1290

10250422 6

= ××× ==

−

Take m = 1291, nm

nL 18.1549

1291

10250422 6

= ××× ==

−

Or take m = 1289, nm

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

8 0

3 years ago