Calculate the linear momentum per photon,energy per photon, and the energy per mole of photons for radiation of wavelength; (a) 600 nm (red), (b) 550 nm (yellow), (c) 400 nm (violet), (d) 200 nm (ultraviolet), (e) 150 pm (X-ray), and (f ) 1.0 cm (microwave).

Answers

Answer 1

Answer:

The detailed explanations is attached below

Explanation:

What is applied is the De brogile equation and the equation showing a relationship between Energy, speed of light and wavelength.

The explanation is as attached below.

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Which of the following is not a risk associated with using legal drugs without medical supervision

Answers

Answer:

paying too much on the black market instead of getting a prescription

Explanation:

i just took the quiz

Answer:

Paying too much on the black market instead of getting a prescription

Explanation:

The rest of the options are risks associated with using legal drugs without medical supervision.

For the system shown below, what is the critical angle (angle at which the system just begins to move)? Assume that the coefficient of friction between all flat surfaces is 0.0500 and that the pulley is frictionless. The mass of m1 is 76.00 kg and the mass of m2 is 194.00 kg. Express your answer in radians.

Answers

THIS IS A PROBLEM OF PHYSICS MECHANIC, PLEASE READ CAREFULLY THE ATTACHED FILE.

Final answer:

To find the critical angle, we need to consider the forces acting on the system. The weight and frictional force must be taken into account. By equating the forces and solving for the critical angle, we can determine at what angle the system just begins to move.

Explanation:

To determine the critical angle for the system shown, we need to consider the forces acting on the objects. The force pulling m1 downwards is its weight, which is equal to its mass multiplied by the acceleration due to gravity. The force preventing m1 from moving is the frictional force, which is equal to the coefficient of friction multiplied by the normal force. The normal force is the force exerted by the surface perpendicular to it, which is equal to the weight of m2 minus the weight of the hanging part of the rope.

At the critical angle, the force of friction is at its maximum value, which is equal to the coefficient of friction multiplied by the normal force. The force pulling m1 downwards is equal to the force of friction. By equating these forces and solving for the critical angle, we can find the answer.

Learn more about critical angle here:

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How many different uniquecurrents will there be in this circuit? How many different "branch" are there in this circuit where the currents could differ.

Answers

Answer:

hello your question is incomplete attached below is the missing circuit diagram

answer : 3 unique currents

Explanation:

From the circuit attached below it can be seen that there will be Three(3) unique currents flowing through this circuit and the Branches where this currents could differ is at the Three(3) resistors

As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

Answers

In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.

The horizontal component of this force is given as

F_x = Fcos(6.7)

While the vertical component of this force would be

F_y = Fsin(6.7)

In the vertical component, the sum of Force indicates that:

$\sum F_y= 0$

The Normal Force would therefore be equivalent to the weight and vertical component of the applied force, therefore:

$N = mg+Fsin(6.7)$

In the horizontal component we have that the Force of tension in its horizontal component is equivalent to the Force of friction:

$\sum F_x = 0$

$F_x = F_(friction)$

$Fcos (6.7) = N\mu$

Using the previously found expression of the Normal Force and replacing it we have to,

$Fcos(6.7)= \mu (mg+Fsin(6.7))$

Replacing,

$Fcos(6.7)= (0.87) (mg+Fsin(6.7))$

$Fcos(6.7) = (0.87)(mg) + (0.87)(Fsin(6.7))$

$Fcos(6.7) -(0.87)(Fsin(6.7)) = 0.87 (mg)$

$F(cos(6.7)-0.87sin(6.7)) = 0.87 (mg)$

$F = (0.87 (mg))/((cos(6.7)-0.87sin(6.7)))$

$F = (0.87(128000*9.8))/((cos(6.7)-0.87sin(6.7)))$

$F = 1.95*10^6N$

Finally the acceleration would be by Newton's second law:

$F = ma$

$a = (F)/(m)$

$a = ( 1.95*10^6)/(128000)$

$a = 15.234m/s^2$

Therefore the greatest acceleration the man can give the airplane is $15.234m/s^2$

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

Answers

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is $\rm 0.430 \; kg\;m^2$ .

Given :

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope.
The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.
The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.
The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.

a) First, determine the acceleration of the B block.

$\rm s = ut + (1)/(2)at^2$

$\rm 1.8 = (1)/(2)* a* (2)^2$

$\rm a = 0.9\; m/sec^2$

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

$\rm \sum F=ma$

$\rm mg-T_b=ma$

$\rm T_b = m(g-a)$

$\rm T_b = 7* (9.8-0.9)$

$\rm T_b = 62.3\;N$

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

$\rm \sum F=ma$

$\rm T_a=ma$

$\rm T_a = 2.1* 0.9$

$\rm T_a = 1.89\;N$

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

$\rm \sum \tau = I\alpha$

$\rm T_br-T_ar = I\alpha$

$\rm I = ((T_b-T_a)r^2)/(a)$

Now, substitute the values of the known terms in the above expression.

$\rm I = ((62.3-1.89)(0.080)^2)/(0.90)$

$\rm I = 0.430 \; kg\;m^2$

For more information, refer to the link given below:

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Answer:

(a) 62.3 N

(b) 1.89 N

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B. There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A. There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?