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Anvisha [2.4K]
3 years ago
5

A transverse wave is found to have a vertical distance of 4 cm from a trough to a crest, a frequency of 12 Hz, and a horizontal

distance of 5 cm from a crest to the nearest trough. Determine the amplitude, period, wavelength and speed of such a wave.
Physics
1 answer:
Alecsey [184]3 years ago
5 0

Answer:

Amplitude: 2cm

Period: 1/12 s

Wavelength: 10 cm

Speed: 120 cm/s

Explanation:

The amplitud is the vertical distance from the resting position to the crest or to the trough, and the vertical distance from a crest to a trough is two times the amplitude. Therefore the amplitud (A) is:

A= 4cm/2= 2cm

The period (T) is:

T=\frac{1}{f}=\frac{1}{12} s

The wavelength is defined as the horizontal distance from a crest to the nearest crest or from a trough to the nearest trough. The distance from a crest to the nearest tough was given. Therefore two times that distance is the distance from a crest to the nearest crest.

The wavelenght (λ) is:

λ=(2)(5)=10cm

The speed (v) in a wave is given by:

v = λf

v=(10)(12)=120 cm/s

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tresset_1 [31]

Answer:

<u>thermal power stations</u>

Explanation:

these resources are burned to produce the electricity.

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2 years ago
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QUESTION 7
ryzh [129]

Answer:

<em>The force required is 3,104 N</em>

Explanation:

<u>Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = ma

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

v_f=v_o+at

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

Solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:

\displaystyle a=\frac{0-20.4}{7.4}

a=-2.757\ m/s^2

The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

F = 1,126\ kg * 2.757\ m/s^2

F= 3,104 N

The force required is 3,104 N

6 0
2 years ago
Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

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Answer:

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Explanation:

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allsm [11]
First, we need to know the amounts of the elements in the compound.

Tin (Sn)= 5.28 g
Fluorine (F) = 8.65 - 5.28 = 3.37 g

Convert these to units of moles by dividing the molar masses.

Tin (Sn)= 5.28 g / 118.71 g/mol = 0.044 mol
Fluorine (F) = 3.37 g / 19.00 g/mol = 0.177 mol

Divide both by the least number of moles of the two.


Tin (Sn)= 0.044 mol /  0.044 mol = 1
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Therefore, the empirical formula would be:
SnF4
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2 years ago
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