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olga nikolaevna [1]
3 years ago
11

Find the current passing through a circuit consisting of a battery and one resistor. The resistor has a resistance of 2 ohms and

the battery has a voltage of 12 volts.
A) 0.167 amps
B) 2 amps
C) 6 amps
D) 8 amps
Physics
2 answers:
-BARSIC- [3]3 years ago
4 0
If one resistor<span> is connected to a battery and has a </span>resistance<span> of 2 </span>ohms<span> and the battery has a </span>voltage<span> of 12 volts, the </span>current<span> passing through the </span>circuit<span> is </span>6 amps. If V=IxR, then I (current) = V (voltage)/ R (resistance<span>) = 12 V/2 </span>Ohms<span> = 6 amps.</span>
I am Lyosha [343]3 years ago
3 0
The current flowing in the circuit can be found by applying Ohm's law:
V=IR
where
V is the voltage of the battery
I is the current flowing in the circuit
R is the resistance of the circuit

In our problem, V=12 V and R=2 \Omega, so if we re-arrange the previous equation and we use these data, we find the current in the circuit:
I= \frac{V}{R}= \frac{12 V}{2 \Omega}=6 A

Therefore the correct answer is C).
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Furkat [3]
The equation for luminous flux is given as P = 4\pir^{2}E
where P is the luminous flux, r is the distance and E is the illumination. The unit for P is lumen, E is lux and r is in meters. Substituting the given to the equation:

P = 4\pir^{2}E 

P= 4\pi(3)^{2}(9.35) = 1057.46 lumens (lm)

The total luminous flux is equal to 1057.46 lumens (lm).
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The density of an object is 1.23 g/cm3. In which of the following materials will it float
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never [62]
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6 0
4 years ago
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¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

7 0
3 years ago
A force vector F1 points due east and has a magnitude of 200N. A second force F2 is added to F1. The resultant of the two vector
PilotLPTM [1.2K]

Answer:

The second vector \vec{F_2} points due West with a magnitude of 600N

Explanation:

The original vector \vec{F_1} points with a magnitude of 200N due east, the Resultant vector \vec{R} points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of  400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:

\vec{F_1}+\vec{F_2}=\vec{R}\implies\vec{F_2}=\vec{R}-\vec{F_1}=-400N-200N=-600N

With the negative sign signifying that the vector points west.

3 0
3 years ago
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