2 years ago

On a guitar if you hit the harder you strike the string the:

Physics

1 answer:

Sedaia [141]2 years ago

6 0

It will get louder A)

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The change in the wind patterns on the earth that causes an increase in ocean temperatures is

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3 years ago

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 n at a

icang [17]

There were 1810 number of people walking.

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3 years ago

A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a

Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = $2\pi$ radians

So, 2.73 revolutions = $2.73\times 2\pi=17.153\ radians$

Therefore, the angular velocity of the tack is, $\omega=17.153\ rad/s$

Now, radial acceleration of the tack is given as:

$a_r=\omega^2 r$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]$

Therefore, the radial acceleration of the tack is 110.9 m/s².

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2 years ago

Electric light was not widely available until the 1930s. True or false

Vlad1618 [11]

Answer:

false is the answer because it was around in the 1880's

Explanation:

I hope this helped

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2 years ago

Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field

morpeh [17]

To develop the problem it is necessary to apply the concepts related to Magnetic Field.

The magnetic field is defined as

$B = \frac{\mu_0 I}{2\pi r}$

Where,

$\mu_0 =$ Permeability constant in free space

r = Radius

I = Current

Our values are given as,

B = 0.1T

d = 4.5mm

r = 2.25mm

If the maximum current that the wire can carry is I, then

$B = \frac{\mu_0 2I}{4\pi r}$

$I = \frac{Br}{2\frac{\mu_0}{4\pi}}$

$I = \frac{(0.1T)(2.25*10^{-3}m)}{2(1*10^{-7}N/A^2)}}$

$I = 1125A$

Therefore the maximum current is 1125A

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3 years ago