PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (40pts)

una onda longitudinal tiene una frecuencia de 200 hz y una longitud de onda de 4.2m ¿cual es la rapidez de la onda?

swat32

Answer:

v = 8.4 m/s

Explanation:

The question ays, "A longitudinal wave has a frequency of 200 Hz and a wavelength of 4.2m. What is the speed of the wave?".

Frequency of a wave, f = 200 Hz

Wavelength = 4.2 cm = 0.042 m

We need to find the speed of the wave. The formula for the speed of a wave is given by :

$v=f\lambda\\\\v=200\times 0.042\\\\=8.4\ m/s$

So, the speed of the wave is equal to 8.4 m/s.

4 0

2 years ago

two 2.5 kg balls move away from each other one traveling 3 m/s to the right the other 4 m/s to the left what is the magnitude of

Juli2301 [7.4K]

Answer:

2.5 kg.m/s

Explanation:

Taking left side as positive while right side direction as negative then

Momentum, p= mv where m is the mass of the object and v is the velocity of travel

Momentum for ball moving towards right side=mv=2.5*-3=-7.5 kg.m/s

Momentum for the ball moving towards the left side=mv=2.5*4=10 kg.m/s

Total momentum=-7.5 kg.m/s+10 kg.m/s=2.5 kg.m/s

5 0

3 years ago

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

$v_{2}=6.28\ m/s$

Put the value into the formula

$P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)$

$\Delta P=1.58\times10^{6}\ N/m^2$

(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

8 0

3 years ago

An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into

Aleksandr-060686 [28]

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

$\rho$ = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N$

The drag force on the athlete is 33 N

3 0

2 years ago

The barometric pressure in breckenridge, colorado (elevation 9600 feet) is 580 mm hg. how many atmospheres is this?

Maksim231197 [3]

1 atmospheric pressure = 760.0 mm Hg

Thus 580 mm Hg = (580 mm Hg/(760 mm Hg/atm))

= 0.763 atm

7 0

2 years ago