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Aleksandr-060686 [28]
3 years ago
8

How could a homework machine help students all around the world?

Physics
1 answer:
slamgirl [31]3 years ago
8 0

Answer:

Hello There!!

Explanation:

The machine can help you alot and do all your homework while you rest and let the homework machine do it and get good grades because of it with no worries but if the machine does it then you won't learn how to do them.

hope this helps,have a great day!!

~Pinky~

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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
A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is
velikii [3]

Answer:

(a)  Length =136.58 m

(b)  T=5995 N

Explanation:

for the glider in the back

T - 1900 = 700 a

for the glider in front

12000-T -1900 = 700a

add equations

12000-3800 = 1400 a

a=5.85 m/s^2

v^2 = v0^2 + 2 a x

40^2 = 2*5.85*x

Length =136.58 m

b) plug the a back into one of the previous formula

T -  1900 = 700*5.85

T=5995 N

8 0
3 years ago
The work-energy theorem states that the work done on an object is equal to a change in which quantity?
Fynjy0 [20]

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

8 0
3 years ago
The second law of thermodynamics says that whenever energy is converted from one form to another in a physical or chemical chang
coldgirl [10]
<span>As per the second law of thermodynamics, when the energy gets converted from one form to another in a physical or chemical change, then the energy which we get as result of change is of lower quality or usability of such energy is less.</span>
3 0
4 years ago
Calculate the amount of energy transferred when a 40w light bulb is left on for 30 minutes.
Alinara [238K]

Answer:

E=72000J or 72kj

Explanation:

The formula is E=pt you need to convert your t from minutes to seconds before proceeding

3 0
3 years ago
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