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2 years ago

How can you use a simple model to describe a wave and its features?

Physics

1 answer:

Volgvan2 years ago

8 0

Answer:

Explanation:

Use mathematical representations to describe a simple model for waves that includes how the amplitude of a wave is related to the energy in a wave. Patterns can be used to identify cause and effect relationships.

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Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

solmaris [256]

Answer:

$\lambda = 3.68 *10^{-36} \ m$

$\lambda_p = 1.28*10^{-14} \ m$

Explanation:

From the question we are told that

The mass of the person is $m = 180 \ kg$

The speed of the person is $v = 1 \ m/s$

The energy of the proton is $E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{h}{m * v }$

Here h is the Planck constant with the value

$h = 6.62607015 * 10^{-34} J \cdot s$

$\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }$

=> $\lambda = 3.68 *10^{-36} \ m$

Generally the energy of the proton is mathematically represented as

$E_p = \frac{1}{2} * m_p * v^2_p$

Here $m_p$ is the mass of proton with value $m_p = 1.67 *10^{-27} \ kg$

=> $8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2$

=> $v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }$

=> $v = 3.09529 *10^{7} \ m/s$

$\lambda_p = \frac{h}{m_p * v_p }$

so $\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }$

=> $\lambda_p = 1.28*10^{-14} \ m$

5 0

3 years ago

A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring balance C and D. Balance C is

vagabundo [1.1K]

A uniform metal rod with of length 80cm and a mass of 3.2kg is supported horizontally by two vertical spring balances C and D. Balance C is 20cm from one end while D is 30cm from the other end would show the reading of 1.06 Kg and 2.13 kg respectively

<h3>What is gravity?</h3>

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another, The gravity varies according to the mass and size of the body for example the force of gravity on the moon is the 1/6th times of the force of gravity on the earth.

As given in the problem, A uniform metal rod of the length of 80cm and mass of 3.2kg is supported horizontally by two vertical springs balance C and D. Balance C is 20cm from one end while D is 30cm from the other end

The weight of the rod acting downward is from the center of the rod at 40 cm

Let us suppose the reading on the spring balance C and D are P and Q respectively

By using the equilibrium for the vertical force

Fv=0

P + C = 3.2

By using the equilibrium for the moment around the left corner

20×P+ 50×Q= 40 ×3.2

By solving for both P and Q from the above two equations we would get

P =1.06 and Q = 2.13

Thus, the reading on the spring balance C and D would be 1.06 Kg and 2.13 kg respectively

Learn more about gravity from here

brainly.com/question/4014727

#SPJ1

5 0

1 year ago

A tank whose bottom is a mirror is filled with water to a depth of 19.6 cm. A small fish floats motionless a distance of 6.40 cm

ANEK [815]

Answer:

A. 4.82 cm

B. 24.66 cm

Explanation:

The depth of water = 19.6 cm

Distance of fish = 6.40 cm

Index of refraction of water = 1.33

(A). Now use the below formula to compute the apparent depth.

$d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.$

(B). the depth of the fish in the mirror.

$d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm$

Now find the depth of reflection of the fish in the bottom of the tank.

$d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8 = 24.66\\$

4 0

3 years ago

Help me please. Please see attached for the questions with the graph

KatRina [158]

Answer

5) b-c

6) a-b and

e-f

7) f-g

9) a-b = 0 m/s

c-d = 0.6667 m/s

e-f = 0 m/s

f-g = -3 m/s

10) b-c ⇒ The cart is acceleration.

e-f ⇒ The cart is moving backwards with a constant velocity.

Explanation

Answer

5) b-c

In the section b-c the cart is accelerating because the slope of the graph is changing. The gradient that represent velocity is increasing.

6) a-b and e-f

At this sections the distance is not changing at all. This can only mean that the cart is not moving. It is at rest.

7) f-g

At this section the slope is negative meaning the cart is moving back to where it came from.

9) a-b = 0 m/s

At a-b the cart is not moving. So the velocity is zero.

c-d = 0.66667 m/s

Velocity = distance / time

=(50-40)/(40-25)

= 10/15

= 0.6667 m/s

e-f = 0 m/s

At e-f the cart is not moving. So the velocity is zero.

f-g = -3 m/s

Velocity = distance / time

= (60-30)/(65-75)

= 30/-10

= - 3 m/s

10) b-c ⇒ The cart is acceleration.

e-f ⇒ The cart is moving backwards with a constant velocity.

3 0

4 years ago

In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or

Aleks04 [339]

Answer:

123 J transfer into the gas

Explanation:

Here we know that 123 J work is done by the gas on its surrounding

So here gas is doing work against external forces

Now for cyclic process we know that

$\Delta U = 0$

so from 1st law of thermodynamics we have

$dQ = W + \Delta U$

$dQ = W$

so work done is same as the heat supplied to the system

So correct answer is

123 J transfer into the gas

4 0

3 years ago