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2 years ago

A proton is located at <3 x 10^-10> m. What is r, the vector from the origin to the location of the proton

Physics

1 answer:

Eduardwww [97]2 years ago

4 0

Complete Question

A proton is located at <3 x 10^{-10}, -5*10^{-10} , -5*10^{-10}> m. What is r, the vector from the origin to the location of the proton

Answer:

The vector position is $\=r=$

Explanation:

From the question we are told that

The position of the proton is $m$

Generally the vector location of the proton is mathematically represented as

$\= r =$

So substituting values

$\=r=$

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3 years ago

Which statement is true of equinoxes? They occur in June and December. Days and nights are equal in length everywhere. The lengt

Zanzabum

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2 years ago

Read 2 more answers

When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point

vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

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

Where,

$v = Velocity \rightarrow 20m/s$

$\lambda = Wavelength \rightarrow 35*10^{-2}m$

Therefore the frequency would be given as

$f = \frac{20}{35*10^{-2}}$

$f = 57.14Hz$

The frequency is directly proportional to the angular velocity therefore

$\omega = 2\pi f$

$\omega = 2\pi *57.14$

$\omega = 359.03rad/s$

Now the maximum speed from the simple harmonic movement is given by

$V_{max} = A\omega$

Where

A = Amplitude

Then replacing,

$V_{max} = (1*10^{-2})(359.03)$

$V_{max} = 3.59m/s$

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8 0

3 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.

4 0

2 years ago

Two cellists, one seated directly behind the other in an orchestra, play the same 220-Hz note for the conductor who is directly

Llana [10]

Answer:

d= 1.56 m

Explanation:

In order to have a constructive interference, the path difference between the sources of the sound, must be equal to an even multiple of the semi-wavelength, as follows:

⇒ d = d₂ - d₁ = 2n*(λ/2)

The minimum possible value for this distance, is when n=1, as it can be seen here:

dmin = λ

In any wave, there exists a fixed relationship between the wave speed, the frequency and the wavelength:

v = λ*f

If v = vsound = 343 m/s, and f = 220 1/s, we can solve for λ:

λ = $\frac{v}{f} = \frac{343 m/s}{220(1/s)} = 1.56 m$

⇒ dmin =λ = 1.56 m

3 0

3 years ago