What will happen in the next moment for the girl on the swing?

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 2.4

NeX [460]

When person is observing destructive interference at 0.20 m distance from the equidistant position then we can say that path difference must be equal to half of the wavelength

now we will have

$\frac{\lambda}{2} = \frac{yd}{L}$

now we know that

y = 0.20 m

d = 2.4 m

L = 10 m

now here we have

$\frac{\lambda}{2} = \frac{0.20\times 2.4}{10}$

$\lambda = 0.096 m$

now frequency of wave is given as

$f = \frac{v}{\lambda}$

$f = \frac{343}{0.096} = 3573 Hz$

5 0

3 years ago

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

or

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0

3 years ago

A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on

klio [65]

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, $q=1.60218\times 10^{-19}\ C$

Mass of a proton, $m=1.67262 \times 10^{-27}\ kg$

The cyclotron frequency is given by :

$f=\dfrac{qB}{2\pi m}$

$f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}$

f = 2286785.40 Hz

or

$\omega=14368296.44\ rad/s$

$\omega=1.43\times 10^7 rad/s$

Hence, this is the required solution.

8 0

2 years ago

which type of organism converts wastes and dead material into nutrients that can be used by plants (1) carnivore (2) herbivore (

Pavlova-9 [17]

Producer hope this helped

7 0

3 years ago

What would happen to the function or purpose of flower if the stigma wasn't sticky?

sashaice [31]

Answer:

The plant would not reproduce because the flower uses the stigma to catch the pollen

8 0

2 years ago

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