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Paladinen [302]
3 years ago
8

What is the value of n?

Physics
2 answers:
vekshin13 years ago
6 0

Answer:

THE ANSWE IS C

81

Explanation:

dusya [7]3 years ago
6 0
81 or 85 I think this because a right angle is 90 degrees and 81 or 85 is a little less
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What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

\Delta P =  \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P is the pressure difference between the two ends

\mu is viscosity of the fluid

L is the length of the pipe

Q=Av is the volumetric flow rate, with A=\pi r^2 being the section of the tube and v the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

\mu=0.001 Pa/s is the dynamic water viscosity at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Using these data in the formula, we get:

\Delta P = 3200 Pa

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0
3 years ago
The sound intensity at a distance 2.00 m from a sound source is 5.00 Find the total sound energy emitted by the source in each s
notka56 [123]

Answer:

     P = 251, 3 W

Explanation:

The intensity is defined as the power emitted per unit area

           I = P / A

Since sound is distributed in all directions spherical shape, the area of ​​a sphere is

           A = 4π r²

let's clear the power and replace

         P = I A

         P = I (4π r²)

let's calculate

         P = 5.00 (4π 2²)

         P = 251, 3 W

6 0
3 years ago
A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-
SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency \omega_0 at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

Where

L = Inductance

C = Capacitance

Our values are given as

C = 15*10^{-6}\mu F

L = 0.280*10^{-3}mH

Replacing we have,

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})

f= 2455.81Hz

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0
3 years ago
6th grade science I mark as brainliest.​
DerKrebs [107]

Answer:

2m 13\frac{1}{3}s

Explanation:

1.5m = 1s

200m = \frac{200}{1.5} × 1s

          = 133\frac{1}{3}s

          = 2m 13\frac{1}{3}s

3 0
3 years ago
14. Which of the following is an example of
Murljashka [212]

Answer:

none of the above or screwdriver

4 0
2 years ago
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