Please help me i need it it’s the only way i’ll pass

A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec

Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec$

Velocity of the string wave is equal to $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec$

Power of wave propagation is equal to $P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt$

So power of the wave will be equal to 5.464 watt

6 0

3 years ago

I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit

Aleks [24]

1) 29.4 N

The force of gravity between two objects is given by:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

$M=5.97\cdot 10^{24} kg$ (mass of the Earth)

$m=3.0 kg$ (mass of the box)

$r=R=6.37\cdot 10^6 m$ (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

$F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N$

2) $g=9.8 m/s^2$

Let's now calculate the ratio F/m. We have:

F = 29.4 N

m = 3.0 kg

Subsituting, we find

$\frac{F}{m}=\frac{29.4}{3.0}=9.8 N/kg = 9.8 m/s^2$

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol $g$ .