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3 years ago

When large amplitude sound vibrates the ear significantly, a _______ results.

2 answers:

7 0

The answer is D. The loudness of a sound, or volume of a sound wave, can be dictated by the abundance of the sound wave. Both pitch and volume can influence how "boisterous" a sound is and how it influences human ears particularly.

kondaur [170]3 years ago

4 0

D loud sound

Hope this helped:)

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Technician a says that the water pump is a centripetal pump. technician b says that centripetal force is the outward force that

Vaselesa [24]

<span>Technician A is correct about the water pump being a centripetal pump and the definition of the centripetal force given by technician B is wrong. The force that happens on a body moving in a circular path and move towards the centre to which the body is moving is called as centripetal force. So itâ€™s not an outward force.</span>

3 0

3 years ago

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

beks73 [17]

Answer:

a. the force between them quadruples

Explanation:

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, the charges on both objects are doubled, so

$q_1' = 2q_1\\q_2' = 2q_2$

While the distance does not change, so the new force will be

$F'= k \frac{(2q_1)(2q_2)}{r^2}=4 (k\frac{q_1 q_2}{r^2})=4 F$

so, the force will quadruple.

4 0

3 years ago

A pushing force acts on a toy car for 3 seconds causing it to accelerate at a rate A’. If the car is replaced with a similar, bu

Alla [95]

Answer:

one-half of A'

Explanation:

3 0

3 years ago

Read 2 more answers

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before

Darina [25.2K]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}$

$\\$

☯ As we know that,

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

$\\$

$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

$\\$

$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

$\\$

$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

$\\$

$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

$\\$

$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

$\\$

☯ For left penetration (s₂)

$\\$

$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

$\\$

$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

$\\$

$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

$\\$

$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

4 0

3 years ago

Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3

Vlada [557]

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

w = 2π f = 2π / T

2π / T = √ k / m

k₀ / m₀ = (2π / T)²

k₀ / m₀ = (2π / 3.0)²

k₀ / m₀ = 4.3865

The period on the new planet is

2π / T = √ k / m

T = 2π √ m / k

In this case the amounts are

m = 6 m₀

k = 10 k₀

We replace

T = 2π√6m₀ / 10k₀

T = 2π √6/10 √m₀ / k₀

T = 2π √ 0.6 √1 / 4.3865

T = 3.23 s

5 0

3 years ago