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inna [77]
1 year ago
6

Max and Jimmy want to jump on a trampoline. Max begins jumping in a steady pattern, making small waves in the trampoline. Jimmy

starts jumping on the trampoline too. Suddenly, Max goes twice as high even though he did not put any extra effort into jumping. What type of wave interaction is this?
Physics
1 answer:
mylen [45]1 year ago
8 0

Answer:

x_total = (A + B) cos (wt + Ф)

we have the sum of the two waves in a phase movement

Explanation:

In this case we can see that the first boy Max when he enters the trampoline and jumps creates a harmonic movement, with a given frequency. When the second boy Jimmy enters the trampoline and begins to jump he also creates a harmonic movement. If the frequency of the two movements is the same and they are in phase we have a resonant process, where the amplitude of the movement increases significantly.

         Max

               x₁ = A cos (wt + Ф)

         Jimmy

              x₂ = B cos (wt + Ф)

         

total movement

             x_total = (A + B) cos (wt + Ф)

 Therefore we have the sum of the two waves in a phase movement

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After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0
2 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
yulyashka [42]

Answer:

178200 g mile pounds

Explanation:

Work= Force * Distance= Fh

F=ma=mg where m is mass and g is acceleration due to gravity

Work= 165 pounds *g* 1080 m=  178200 g mile pounds

5 0
2 years ago
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl
tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

5 0
1 year ago
A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the ab
belka [17]

Answer:

A)   θ = 13.1º  , B)  E

Explanation:

A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.

In this reference system the only force that we must decompose is the Normal one, let's use trigonometry

        sin θ = Nₓ / N

        cos θ = N_{y} / N

        Nₓ = N sin θ

       Ny = N cos θ

x-axis (radial)

        Nₓ = m a

where the acceleration is centripetal

         a = v² / R

we substitute

        -N sin θ = -m v² / R                   (1)

the negative sign indicates that the force and acceleration towards the center of the circle

y-axis (Vertical)

          Ny - W = 0

           N cos θ = mg

           N = mg / cos θ

we substitute in 1

          mg / cos θ  sin θ = m v² / R

          g tan θ = v² / R

          θ = tan⁻¹ (v² / gR)

we calculate

        θ = tan⁻¹ (25² / 9.8 274)

        θ = 13.1º

B) when comparing the equations the correct one is E

6 0
1 year ago
A 100-kg box is sitting on a 10 degree incline with a coefficient of friction of 0.5. At what angle must the incline be raised t
Marta_Voda [28]

Answer:

26.56°

Explanation:

μ= 0.5

F = ?

N =?

Fs = μN

Fs = force acting due to friction (on an inclined plane)

N = normal force.

μ= coefficient of friction

μ= tanΘ

μ= tan 0.5

Θ = tan⁻¹ 0.5

Θ = 26.56°

The angle in which is required for the box to start moving across the inclined plane is 26.56°

5 0
1 year ago
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