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Tanya [424]
2 years ago
14

T or F? An electric field has strength, but no direction.????

Physics
1 answer:
Aleks04 [339]2 years ago
5 0

Answer:

FALSE

Explanation:

an electric field has both magnitude and direction. it also has strength

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Y_Kistochka [10]
Important bc it makes it more effective, the specific rate makes or breaks the fitness. Frequency is important to allow your body to rebuild and repair the damage from working out, it allows the body to adapt and time for rest/ healing. Intensity depends on how much your body breaks so the recover time and frequency must be adjusted. Time effects because of the distance between frequencies which plays a role.
5 0
3 years ago
A 16.0 Ω, 13.0 Ω, and 7.00 Ω resistor are connected in parallel to an emf source. A current of 6.00 A is in the 13.0 Ω resistor.
Darya [45]

Answer:3.54ohms

Explanation: connection in parallel

1/Rt= 1/R1+1/R2+1/R3

1/Rt= 1/16+1/13+1/7

1/Rt= 91+112+208/1456

1/Rt= 411/1456

411Rt= 1456

Rt= 1456/411

Rt= 3.54ohms

3 0
4 years ago
Read 2 more answers
Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge
kirill [66]

Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The  y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

f) The sum of both x components gives :

Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²\sqrt{((17.4)^{2} +(16.9)^{2}} = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

4 0
3 years ago
True or False?
WINSTONCH [101]

Answer:

true is the answer of the question

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3 years ago
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What is the symbol for ammonium
rosijanka [135]
Symbol of ammonium is

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