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3 years ago

12

Explain what happens to the motion of a particle as a wave passes through a medium. How is the motion of the particles like the

motion of a mass on a spring

1 answer:

bezimeni [28]3 years ago

3 0

As they vibrate, they pass the disturbance of energy to the particles next to them, which passes the energy to the particles next to them, and so on.
Energy travels down the spring and reflected like the motion of a mass on a spring.

<u>Explanation</u>:

Particle motion is the superposition of a large component due to the fluid drag toward the filter barrier and a random component due to Brownian motion.

The particles of the medium just vibrate in place. As they vibrate, they pass the unsettling influence of energy to the particles beside them, which passes the energy to the particles by them, etc.

Energy goes down the spring and reflected like the movement of a mass on a spring.

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Select the correct answer.

chubhunter [2.5K]

I believe it’s self-referent encoding

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2 years ago

Describe J.J Thomson's plum pudding model,and explain how it compares to the modern day atomic theory.

juin [17]

The correct answer to this open question is the following.

The J.J Thomson's plum pudding model establishes that when he researched the atom, he found that electrons are positively charged. Because he observed that electrons were embedded like plums. Electrons are negative. However, compared to the modern-day atomic theory, scientist Ernest Rutherford established that the space between an atom is empty, the nucleus of the atom is positive, and negative electrons have fixed motion paths.

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2 years ago

Read 2 more answers

I need help please I will give points

stira [4]

Answer:

-5 N of force

Explanation:

Hope it helps!

3 0

2 years ago

Question 1 of 10

Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field $E$ at a distance $R$ from a charge $Q$ is given by

$E = k\dfrac{Q}{R^2}$

where $k = 9*10^9Nm/C$ is the coulomb's constant.

Now, in our case

$R = 0.0075m$

$Q = 0.0035C$ ;

therefore,

$E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}$

$\boxed{E = 5.6*10^{11}N/C.}$

which is choice C from the options given<em> (at least it resembles it).</em>

6 0

2 years ago

Find the current flowing out of the battery.

klemol [59]

Answer:

0.36 A.

Explanation:

We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:

Resistor 1 (R₁) = 35 Ω

Resistor 2 (R₂) = 20 Ω

Equivalent Resistance (Rₑq) =?

Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:

Rₑq = (R₁ × R₂) / (R₁ + R₂)

Rₑq = (35 × 20) / (35 + 20)

Rₑq = 700 / 55

Rₑq = 12.73 Ω

Next, we shall determine the total resistance in the circuit. This can be obtained as follow:

Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω

Resistor 3 (R₃) = 15 Ω

Total resistance (R) in the circuit =?

R = Rₑq + R₃ (they are in series connection)

R = 12.73 + 15

R = 27.73 Ω

Finally, we shall determine the current. This can be obtained as follow:

Total resistance (R) = 27.73 Ω

Voltage (V) = 10 V

Current (I) =?

V = IR

10 = I × 27.73

Divide both side by 27.73

I = 10 / 27.73

I = 0.36 A

Therefore, the current is 0.36 A.

6 0

3 years ago