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maria [59]
2 years ago
7

8. Find the momentum of a photon in eV/c and in Kg. m/s if the wavelength is (a) 400nm ; (b) 1 Å = 0.1 nm, (c) 3 cm ; and (d) 2

nm .
Physics
1 answer:
nataly862011 [7]2 years ago
7 0
We use the formula: p = E/c where E = hc / λ. hence, p = h/  λ. where h is the Planck's constant: 6.62607004 × 10-34 m2 kg / s and <span>λ is the wavelenght. 
</span>
a) p = <span>6.62607004 × 10-34 m2 kg / s / 0.1 x10^-9 m = 6.62607 x 10-24 m kg/s
</span>b) p = 6.62607004 × 10-34 m2 kg / s / 3 x10^-2 m = 2.20869 <span>x 10-32 m kg/s
</span>b) p = 6.62607004 × 10-34 m2 kg / s / 2 x10^-9 m = 3.3130 <span>x 10-25 m kg/s</span>
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A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in
Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
  • \boxed{a = acceleration \ (constant)}
  • \boxed{d = distance \ travelled}

The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
  • v = final velocity
  • d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>
  1. Determine the acceleration of the stuffed bear brainly.com/question/6268248
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Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0
2 years ago
Read 2 more answers
Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.
Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 1 (air)                  E = 7.5 / 0.10*10^-3

                                E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

                                C = k*A*ε / d

k = 1 (air)                   C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 2.213 pF

                                Q = C*V

                                Q = 7.5*(2.213)

                                Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?    

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 3.13                     E = 7.5 / (3.13)0.10*10^-3

                                E = 24,000 V/m              

F) What is the capacitor's charge after the Mylar is inserted?      

                                C = k*A*ε / d

k = 3.13                    C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 6.927 pF

                                Q = C*V

                                Q = 7.5*(6.927)

                                Q = 52 pC                                      

6 0
2 years ago
An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall a
blsea [12.9K]

Answer: apparent weighlessness.


Explanation:


1) Balance of forces on a person falling:


i) To answer this question we will deal with the assumption of non-drag force (abscence of air).


ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).


iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).


2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.


3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.


Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

3 0
2 years ago
Read 2 more answers
A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block
solniwko [45]
We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
4 0
2 years ago
A uniform magnetic field makes an angle of 30o with the z axis. If the magnetic flux through a 1.0 m2 portion of the xy plane is
Irina18 [472]

Answer:

(b) 10 Wb

Explanation:

Given;

angle of inclination of magnetic field, θ = 30°

initial area of the plane, A₁ = 1 m²

initial magnetic flux through the plane, Φ₁ =  5.0 Wb

Magnetic flux is given as;

Φ = BACosθ

where;

B is the strength of magnetic field

A is the area of the plane

θ is the angle of inclination

Φ₁ = BA₁Cosθ

5 = B(1 x cos30)

B = 5/(cos30)

B = 5.7735 T

Now calculate the magnetic flux through a 2.0 m² portion of the same plane

Φ₂ = BA₂Cosθ

Φ₂ = 5.7735 x 2 x cos30

Φ₂ = 10 Wb

Therefore, the magnetic flux through a 2.0 m² portion of the same plane is is 10 Wb.

Option "b"

3 0
2 years ago
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