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3 years ago

A TV draws 0.867 A of current when connected to a 122 V outlet. It’s resistance is 140.71.

Physics

1 answer:

Rina8888 [55]3 years ago

8 0

Answer:

Explanation:

current measures charge flowing per second .

A current of .867 A means charge of .867 coulomb is flowing per second

charge flowing per minute = .867 x 60

= 52.02 coulomb

one electrons carries a charge of 1.6 x 10⁻¹⁹ Coulomb

So no of charge flowing per minute

= 52.02 / 1.6 x 10⁻¹⁹

= 32.5 x 10¹⁹ .

B )

electrical energy consumed per minute

= V X I X t , V is volt , I is current and t is time .

= 122 x .867 x 60

= 6346.44 J .

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A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starti

Zigmanuir [339]

Answer:

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

From the free fall movement we have the following formulas: $Vf^{2} = Vo^{2} - 2gh$ and $h=Vo*t - \frac{g*t^{2} }{2}$ , First we need to find the height to time iqual to 3.04 s using the formula: $h=Vo*t - \frac{g*t^{2} }{2}$ and remember that golf ball was released from the rest (Vo= 0 (m/s)) so $h= (0 (m/s))*(3.04 (s)) - \frac{9.8 (m/s^2)*(3.04 (s))^{2} }{2}$ , we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using $Vf^{2} = Vo^{2} - 2gh$ solving for Vf, we get: $Vf = \sqrt{Vo^{2}-2*g*h }$ replacing the values given $Vf = \sqrt{(0 m/s)^{2}-2*(9.8 m/s^2)*(-45.28 m) }$ , so we get: Vf = 29.79 (m/s).

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3 years ago

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Which statement describes a surface wave?

nexus9112 [7]

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Which of the following is/are true? Electromagnetic waves are created by accelerating charges. Electromagnetic waves are transve

Vikki [24]

Answer:

All of the above are true.

Explanation:

(a). true

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(b). true

the electric field and the magnetic field have vibrations in the perpendicular direction along the motion of the wave so electromagnetic wave is a transverse wave. therefore, the EM wave is a Transverse wave

(c) true .

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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.

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