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2 years ago

A diffraction grating having 500 lines/mm diffracts visible light at 30°. What is the light’s wavelength?

Physics

1 answer:

makvit [3.9K]2 years ago

8 0

Answer:

The light’s wavelength is 500 nm

Explanation:

mλ = d*sin(θm)

where;

λ is the visible light wavelength (nm) = ?

d is the groove (slit) spacing (mm/lines) = 1/500mm

θm is the angle of refraction

m is is the diffraction order, for visible light range, m = 2

λ = [d*sin(θm)]/m

λ = (sin30)/(500*2) = (0.5)/(1000) = 0.0005 mm = 500 nm

Therefore, for a diffraction grating having 500 lines/mm and diffracts visible light at 30°. The light’s wavelength is 500 nm

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Determine the sign (+ or −) of the torque about the elbow caused by the biceps, τbiceps, the sign of the weight of the forearm,

Alex Ar [27]

Ans:
1. τbiceps = +(Positive)
2. τforearm = -(Negative)
3. τball = -(Negative)

Explanation:

The figure is attached down below.

1. Torque about the elbow caused by the biceps, τbiceps:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the biceps.

We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

3. Torque about the the weight of the ball, τball:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the ball.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(ball) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

8 0

2 years ago

A spaceship of mass 8600 kg is returning to Earth with its engine turned off. Consider only the gravitational field of Earth. Le

Katyanochek1 [597]

Answer:

$\Delta KE = 4.20\times 10^{13}\ J$

Explanation:

given,

mass of spaceship(m) = 8600 Kg

Mass of earth = 5.972 x 10²⁴ Kg

position of movement of space ship

R₁ = 7300 Km

R₂ = 6700 Km

the kinetic energy of the spaceship increases by = ?

Increase in Kinetic energy = decrease in potential energy

$\Delta KE = GMm (\dfrac{1}{R_2}-\dfrac{1}{R_1})$

$\Delta KE = GMm (\dfrac{R_1-R_2}{R_2R_1})$

$\Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{7300 - 6700}{7300 \times 6700})$

$\Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{600}{48910000})$

$\Delta KE = 4.20\times 10^{13}\ J$

5 0

2 years ago

A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet

kogti [31]

The centripetal force, Fc, is calculated through the equation,
Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius.
Substituting the known values,
Fc = (112 kg)(8.9 m/s)² / (15.5 m)
= 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N.

4 0

2 years ago

Read 2 more answers

Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)

Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force, $(FS_p)s=-k\Delta s-q(\Delta s)^3$

$k=350N/m$

$q=750 N/m^3$

We have to find the work must you do to compress this spring 15 cm.

$\Delta s=15 cm=0.15 m$

Using 1 m=100 cm

Work done= $\int_{0}^{0.15}-Fd(\Delta s)$

W= $-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)$

$W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}$

$W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750$

$W=4.033 J$

Ideal spring work= $0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J$

Percentage increase in work= $\frac{4.033-3.938}{3.928}\times 100=2.4$ %

6 0

2 years ago

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

Virty [35]

Answer:

1. The tension in the rope is everywhere the same.

2. The magnitudes of the forces exerted on the two objects by the rope are the same.

3. The forces exerted on the two objects by the rope must be in opposite directions.

Explanation:

"Massless ropes" do not have a "net force" which means that it is able to transmit the force from one end of the rope to the other end, perfectly. It is known for its property of having a total force of zero. In order to attain this property, the magnitude of the forces exerted on the two stationary objects by the rope are the same and in opposite direction. So this explains number 2 & 3 answers.

Since the objects that are held by the rope are stationary, then this means that the tension in the rope is also stationary. This means that the tension in the rope everywhere is the same (provided that the rope is still or in a straight line, as stated in the situation above, and is being held by two points). So, this explains number 1.

6 0

2 years ago