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disa [49]
2 years ago
15

How do I solve this?​

Physics
1 answer:
prohojiy [21]2 years ago
3 0

unlock your inner brain

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One problem when using rigid metal conduit in a residence is that the installation of the conduit may a weaken the structure b r
worty [1.4K]
The correct answer is A. Installation of rigid metal conduit requires grounding and the grounding equipment used may weaken the structure.
8 0
3 years ago
Read 2 more answers
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
3 years ago
Radio waves transmitted through space at 3.00 ✕ 108 m/s by the Voyager spacecraft have a wavelength of 0.137 m. What is their fr
fredd [130]

Answer:

Frequency, f=2.18\times 10^9\ Hz

Explanation:

We have,

Speed of radio waves is 3\times 10^8\ m/s

Wavelength of radio waves is \lambda=0.137\ m

It is required to find the frequency of the radio waves. The speed of a wave is given by :

v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.137}\\\\f=2.18\times 10^9\ Hz

So, the frequency of the radio wave is 2.18\times 10^9\ Hz.

8 0
3 years ago
An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of
GaryK [48]

Explanation:

The given data is as follows.

     Velocity of bullet, c_{p} = 814.8 m/s

    Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

               t = \frac{24.7}{343}      (velocity in air = 343 m/s)

                 = 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

               d_{b} = c_{b} \times t

                          = 814.8 \times 0.072

                          = 58.66

                          = 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0
3 years ago
A 50.0-g Super Ball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A
Setler [38]

Explanation:

Average acceleration is change in velocity over time.

a = Δv / Δt

a = (22.0 m/s − (-25.0 m/s)) / 0.00350 s

a = 13,400 m/s²

7 0
3 years ago
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