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2 years ago

Write a use of reverberation

Physics

1 answer:

MakcuM [25]2 years ago

3 0

Hello there!

definition of reverberation: "Prolongation of a sound; resonance."

reverberation in a sentence: The school's special effects team added reverberation to the actress's monologue to make her speech more dramatic and the school play a little more interesting.

I hope this helps, have a great day! ^-^

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As more and more resistors are added in parallel to a circuit, the equivalent resistance of the circuit ____________ (increases,

Anestetic [448]

Answer:

Decreases, Increases

Explanation:

Resistance is parallel can be calculated using

1/Req = 1/R1 + 1/R2 + 1/R3 +....

Then, as more resistor is added in parallel the equivalent resistance is reduced.

Let use a simple sample

Let all the resistor have equal resistances

Let say R = R1 = R2 = R3 =...Rn

Then, 1/Req = 1/R1 + 1/R2 + 1/R3 +....

1/Req = 1/R + 1/R + 1/R +.... 1/Rn

Req = R/n

Check attachment on how I got that.

This implies that, the equivalent resistance will always be less than the original resistance, since n>1

So, as n increases (I.e. as the number of resistance increases), the equivalent resistance reduces.

B. Now, to know if the current reduces or increases

Using Ohms law

V = iR

Then, I = V/R

So, let assume the voltage is constant, then, the current is inversely proportional to the resistance, so as we know that the resistance is reducing, then the current will be increasing.

So current increase as we add more resistor in parallel to a circuit

3 0

2 years ago

What would be the answer for this and how?

Veronika [31]

Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

$k = \frac{F}{\Delta x}\\$

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

Therefore,

$k = \frac{10\ N}{0.04\ m}\\k = 250\ N/m\\$

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

$k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\$

For a load of 30 N, applying Hooke's Law:

$\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\$

Hence, the correct option is:

B. 6 cm

7 0

2 years ago

Cozebilir misiniz !!!

GREYUIT [131]

Answer:

Ya inan bana çok özür dilerim cevabı bilmiyorum ama ilk defa bugün indirdim ve hiç türk yoktu selam vermek istedim selam jwlwpskwks herkes yabancı burda neden?

5 0

2 years ago

Modern nuclear bomb tests have created an extra high level of14C in our atmosphere. When future archaeologistsdate samples from

Ipatiy [6.2K]

Answer:

Too old(Ex. if real time is 1000 then they estimate >1000)

Explanation:

This is because with time our planet may have a definite function which describes temperature.(Because of all the factors and global warming except nuclear bomb testing)

Now nuclear test on planet have significant effect on temperature rise.

Also 14°C rise in temperature is good one because of this.

If future archaeologists only consider that uniform function as above mentioned then they estimate more time then the real one.

Thus too old is right answer.

8 0

2 years ago

The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro

qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So, Q = ∫∫∫ρdV

Q = ∫∫∫ρr²sinθdθdrdΦ

Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q = ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ

Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q = ∫∫0.2r⁴[-cosθ]drdΦ

Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q = ∫∫0.2r⁴-[- 2]drdΦ

Q = ∫∫0.2r⁴(2)drdΦ

Q = ∫∫0.4r⁴drdΦ

Q = ∫0.4r⁴dr∫dΦ

Q = ∫0.4r⁴dr[Φ]

Q = ∫0.4r⁴dr[2π - 0]

Q = ∫0.4r⁴dr[2π]

Q = ∫0.8πr⁴dr

Q = 0.8π∫r⁴dr

Q = 0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0

2 years ago