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3 years ago

How will decreasing the amplitude of the sound waves from a television affect its intensity?

Physics

1 answer:

Colt1911 [192]3 years ago

6 0

A. The sound will decrease in volume

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In which era did the universe’s clouds start to condense and the universe became transparent for the first time?

alukav5142 [94]

Answer:

The Universe became transparent to the light left over from the Big Bang when it was roughly 380,000 years old

Explanation:

8 0

3 years ago

Describe how the mass, luminosity, surface temperature, and radius of main-sequence stars change in value going from the “bottom

Sladkaya [172]

Answer:

1. Least massive stars are the coolest and least luminous, lower right of main sequence, on HR diagram.

2. Most massive are the hottest and most luminous, upper left of main sequence on Hr Diagram.

3. The radius of stars are related to their sprectral type. having the O being the hottest upper left and M being the coolest bottom right.

4 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

2 years ago

How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s

DIA [1.3K]

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

a = 3 m/s^2

Formula

a = (vf - vi) /t Substitute the givens into this formuls

Solution

3 = (40 - 10) / t Multiply both sides by t

3*t = t(40 - 10)/t Combine. Cancel t's on the right

3*t = 30 Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

6 0

2 years ago

A toroid has a square cross section with the length of an edge equal to the radius of the inner surface. The ratio of the magnit

Tasya [4]

Answer:

Explanation:

7 0

3 years ago

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