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satela [25.4K]
3 years ago
6

An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t

his field through a square of surface area A = 2 m2 oriented parallel to the yz-plane?
Physics
1 answer:
Ede4ka [16]3 years ago
4 0

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0

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A student is considering doing a complete repeated measures design experiment involving motor skills. The student's advisor has
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Answer:

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If participants experience conditions more than once, they experience the conditions first in one order, then the opposite order.

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Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr
IgorLugansk [536]

(a) 273.9 V

The power rating of the resistor is given by

P=\frac{V^2}{R}

where

P is the power rating

V is the potential difference across the resistor

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If the maximum power rating is P=5.0 W, and the resistance of the resistor is R=15 k\Omega = 15000 \Omega, then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V:

V=\sqrt{PR}=\sqrt{(5.0 W)(15000 \Omega)}=273.9 V

(b) 1.6 W

In this case, we have:

R=9.0 k\Omega = 9000 \Omega is the resistance of the resistor

V=120 V is the potential difference across the resistor

So we can find the power rating by using the same formula of part (a):

P=\frac{V^2}{R}=\frac{(120 V)^2}{9000 \Omega}=1.6 W

(c) Maximum voltage: 14.1 V; Rate of heat: 2.00 W and 3.00 W

Here we have two resistors of

R_1 = 100 \Omega\\R_2 = 150 \Omega

and each resistor has a power rating of

P = 2.00 W

So the greatest potential difference allowed in the first resistor is

V=\sqrt{PR_1}=\sqrt{(2.00 W)(100 \Omega)}=14.1 V

While the greatest potential difference allowed in the second resistor is

V=\sqrt{PR_2}=\sqrt{(2.00 W)(150 \Omega)}=17.3 V

So the greatest potential difference allowed not to overheat either of the resistor is 14.1 V.

In this condition, the power dissipated on the first resistor is 2.00 W, while the power dissipated on the second resistor is

P_2 = \frac{V^2}{R_2}=\frac{(14.1 V)^2}{150 \Omega}=1.33 W

And this corresponds to the rate of heat generated in the first resistor (2.00 W) and in the second resistor (1.33 W).

4 0
2 years ago
A proton moves in the negative x-direction through a uniform magnetic field in the negative y-direction what is the direction of
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A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction. The proton will be subject to a magnetic pull that is directed into the page. Option B is correct.

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A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction.

The proton will be subject to a magnetic pull that is directed into the page.

Hence, option B is correct.

To learn more about the right-hand thumb rule refer to the link;

brainly.com/question/11521829

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2 years ago
VERY EASY QUESTION!!!!!! 20 POINTS!!!!!!!!
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