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3 years ago

A 6-ohm resistor and a 4-ohm resistor are

Physics

1 answer:

SpyIntel [72]3 years ago

6 0

The current flowing in the circuit is given by the Ohm's First Law.

$i_{eq}= \frac{V}{R_{eq}} \\ i_{eq}= \frac{6}{4+6} \\ i_{eq}= 0.6A$

The current flowing in the 6-ohm resistor is the total current of circuit, therefore the voltage drop across this resistor is given by:

$V_{6}=R_{6}*I_{eq} \\ V_{6}=6*0.6 \\ \boxed {V_{6}=3.6V}$

Obs: the drawing is attached

If you notice any mistake in my english, please let me know, because i am not native.

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A rock is thrown vertically upward from the origin of a reference frame. The rock reaches the peak of its trajectory and then it

Tema [17]

Answer:

The acceleration of the rock is negative throughout the whole motion. (TRUE)

The acceleration of the rock is zero at the peak of the trajectory (FALSE)

The velocity of the rock is positive throughout the whole motion (FALSE)

The position of the rock is positive or zero throughout the whole motion (TRUE)

The velocity of the rock is positive throughout the whole motion (FALSE)

Explanation:

1)The acceleration of the rock is negative throughout the whole motion. (TRUE)

Using the upward direction as positive

Then, The acceleration of the rock is negative throughout the whole motion.

2)The acceleration of the rock is zero at the peak of the trajectory.(FALSE)

Only the velocity can be zero at the peak the trajectory, the acceleration is not zero

3)The velocity of the rock is positive throughout the whole motion (FALSE)

Using the upward direction as positive, The velocity of the rock is only positive when going in upward direction and vice versa

4)The position of the rock is positive or zero throughout the whole motion (TRUE)

When the rock get down it will have a zero position, then when it get to the peak it will have positive position

6 0

4 years ago

A LASIK vision-correction system uses a laser that emits 10-ns-long pulses of light, each with 3.0 mJ of energy. The laser beam

lyudmila [28]

<span>P = energy/t = 0.0025/1E-8 = 250000 W
I(ave) = P/A = 250000/(pi*0.425E-3^2) = 4.4056732E11 W/m^2
I(peak) = 2I(ave) = 8.8113463E11 W/m^2
Electric field E = sqrt(I(peak)*Z0) = 1.8219499E7 V/m, where
free-space impedance Z0 = sqrt(µ0/e0) = 376.73031 ohms</span>

7 0

3 years ago

Based on its type of chemical bond, which of the following has the highest boiling point?

Sever21 [200]

<h3><u>Answer;</u></h3>

Potassium chloride

<h3><u>Explanation;</u></h3>

Potassium Chloride is an ionic compound and has ionic bond which is stronger than covalent bond in ethyl alcohol,water ,ammonia, and thus has the highest boiling point .
ionic bond is a type of bond that results from the transfer of electrons between metallic atoms and non-metallic atoms.

5 0

3 years ago

Frequency of the wave below?

agasfer [191]

It's hard to tell exactly what's happening in that 110 cm that you marked over the wave. What is under the ends of the long arrow ? How many complete waves ? I counted 4.5 complete waves ... maybe ?

If there are 4.5 complete waves in 110cm, then the length of 1 wave is (110/4.5)=24.44cm.

Frequency = speed/wavelength

Frequency = 2m/s /0.2444m

Frequency = 8.18 Hz

6 0

3 years ago

A soccer goal is 2.44 m high. A player kicks the ball at a distance 13 m from the goal at an angle of 40°, and the ball just hit

Vera_Pavlovna [14]

Let <em>v</em> denote the initial speed of the ball. The ball's position at time <em>t</em> is given by the vector

$\mathbf r(t)=v\cos40^\circ\,t\,\mathbf i+\left(v\sin40^\circ\,t-\dfrac g2t^2\right)\,\mathbf j$

where <em>g</em> is the acceleration due to gravity with magnitude 9.80 m/s^2.

The ball reaches the goal 13 m away at time <em>t</em> such that

$10\,\mathrm m=v\cos40^\circ t\implies t=\dfrac{10\,\mathrm m}{v\cos40^\circ}$

at which point it attains a height of 2.44 m, so that

$2.44\,\mathrm m=v\sin40^\circ\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)-\dfrac g2\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)^2$

$2.44\,\mathrm m=(10\,\mathrm m)\tan40^\circ-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)\left(\dfrac{100\,\mathrm m^2}{v^2\cos^240^\circ}\right)$

$\implies\boxed{v\approx3.75\dfrac{\rm m}{\rm s}}$

6 0

4 years ago