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earnstyle [38]
3 years ago
6

What factors affect the speed of water waves

Physics
1 answer:
Aneli [31]3 years ago
6 0
Hey there,

Your question states: What factors affect the speed of water waves
Let's get one thing out the way, (wavelength) does NOT affect the the speed of water. If anything, it would be how high the wavelength's are. The higher the wavelengths are, the more that it would affect the speed, because there very high, but if it were to go longer on the width side, that would increase the speed, but that's not the case. Your correct answer would be (higher wavelength).

Hope this really helps you.
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A series RLC circuit with L = 12 mH, C = 3.5 mu or micro FF, and R = 3.3 ohm is driven by a generator with a maximum emf of 115
Elan Coil [88]

Explanation:

Given data

Inductance L=12*10^-³H

Capacitance C= 3.5*10^-6F

Resistance R= 3.3 Ohms

Voltage V=115v

Capacitive reactance Xc=?

inductive reactance Xl=?

Impedance Z=?

Phase angle =?

A. Resonance frequency

In RLC circuit resonance occurs when capacitive reactance equals inductive reactance

f=1/2pi √ LC

f=1/2*3.142 √ 12*10^-³*3.5*10^-6

f=1/6.284*0.0002

f=1/0.00125

f=800HZ

B. Find Irms at resonance.

Irms=R/V

Irms=3.3/115

Irms=0.028amp

Find the capacitive reactance XC in Ohms

Xc=1/2pi*f*C

Xc=1/2*3.142*800*3.5*10^-6

Xc=1/0.0176

Xc=56.8 ohms

To find the inductive reactance

Xl=2pifL

Xl=2*3.142*800*12*10^-3

Xl=60.3ohms

d) Find the impedance Z.

Z=√R²+(Xl-Xc)²

Z=√3.3²+(60.3-56.8)²

Z=√10.89+12.25

Z=√23.14

Z=4.8ohms

Phase angle =

Tan phi=Xc/R=56.8/3.3

Tan phi=17.2

Phi=tan-1 17.2

Phi= 1.51°

6 0
3 years ago
Analyze the image below and answer the question that follows.
goldenfox [79]
There are many principles that are classified under the Gestalt Principles. The Gestalt principles believe that any stimulus can be viewed in a very simple form. According to the given image, I can say that the square represents the Gestalt principle of PROXIMITY. Hope this helps.
7 0
3 years ago
wich of the following are commonly distributed by veterinary assistants in typical veterinary practice?
vredina [299]
Can you sent a picture of the answers
5 0
3 years ago
A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a
tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is  suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}

W = 3.266 N

The mass of the meters stick is :

m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg

So, the mass of the meter stick is 0.333 kg.

5 0
3 years ago
Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

4 0
3 years ago
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