Someone please help me!!! Please explain and I will mark brainliest!!!

A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels

katovenus [111]

To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

$v = \sqrt{\frac{T}{\mu}}$

Here,

v = Velocity

$\mu$ = Linear density (Mass per unit length)

T = Tension

Rearranging to find the Period we have that

$T = v^2 \mu$

$T = v^2 (\frac{m}{L})$

As we know that speed is equivalent to displacement in a unit of time, we will have to

$T = (\frac{L}{t}) ^2(\frac{m}{L})$

$T = (\frac{7.8}{0.83})^2 (\frac{0.49}{7.8})$

$T = 5.54N$

Therefore the tension is 5.54N

8 0

3 years ago

Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with

Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is $2.728\times 10^{-3}\,\frac{m}{s^{2}}$ ( $\frac{2.784}{10000}\cdot g$ , where $g = 9.8\,\frac{m}{s^{2}}$ ).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration ( $g$ ), in meters per square second, is directly proportional to the mass of the Earth ( $M$ ), in kilograms, and inversely proportional to the distance from the center of the Earth ( $r$ ), in meters:

$g = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$G$ - Gravitational constant, in cubic meters per kilogram-square second.

$M$ - Mass of the Earth, in kilograms.

$r$ - Distance from the center of the Earth, in meters.

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972\times 10^{24}\,kg$ and $r = 382.26\times 10^{6}\,m$ , then the free-fall acceleration at the orbit of the Moon is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}$

$g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}$

6 0

3 years ago

Water is most dense at 4 degrees Celsius. Since at this temperature 1 ml of water has a mass of 1 g. What is its density?

sveticcg [70]

The density is 1. 1/1=1

7 0

3 years ago

Read 2 more answers

Shorter the vibrating part more will be the pitch. How?

Lina20 [59]

Answer:When the length of a string is changed, it will vibrate with a different frequency.Shorter strings have higher frequency and therefore higher pitch.

6 0

3 years ago

Suppose the electric field between two parallel plates kept constant but the distance between them is tripled. What happens to t

alukav5142 [94]

Answer:

The voltage increases by a factor of three.

Explanation:

7 0

3 years ago