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Zigmanuir [339]
3 years ago
14

If the current in the circuit above is 10 amps and the resistance is 3 ohms, what is the voltage?

Physics
1 answer:
Naddik [55]3 years ago
4 0
It's really difficult to make out the circuit above.  Quite frankly, your
question leaves me to wonder how far 'above' it may be. 
The best I can do will be to try and fabricate an answer based on the
information given in the text of the question, augmented only by my own
training, chutzpah, and life experiences.

If the circuit ... wherever it is ... consists entirely of the single 3-ohm
resistance and no other components, and the current through the
resistance is 10 Amperes, then

             Voltage = (Current) x (resistance)  =  30 volts .
You might be interested in
In this circuit (see picture), which resistor will draw the least power?
Basile [38]
A few different ways to do this: 

Way #1: 
The current in the series loop is  (12 V) / (total resistance) . 
(Turns out to be 2 Amperes, but the question isn't asking for that.)

In a series loop, the current is the same at every point, so it's
the same current through each resistor.

The power dissipated by a resistor is  (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power.  That's  R1 .

And by the way, it's not "drawing" the most power.  It's dissipating it.

Way #2:
Another expression for the power dissipated by a resistance is

                 (voltage across the resistance)²  /  (resistance)  .

In a series loop, the voltage across each resistor is

          [ (individual resistance) / (total resistance ] x battery voltage.

So the power dissipated by each resistor is

         (individual resistance)² x [(battery voltage) / (total resistance)²]

This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .                                      

Way #3:  (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.

===>  When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).
3 0
3 years ago
If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll
faltersainse [42]
<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

a) Given S(t) = 76 + 128t − 16t²

    s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    Displacement in 4 seconds = 332 - 76 = 256 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{256}{4}=64ft/s

    Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

    Displacement in 4 seconds = 78 - 332 = -254 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{-254}{4}=-63.5ft/s

    Average velocity in second 4 seconds is 63.5 ft/s downward

3 0
3 years ago
Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
Nataliya [291]

Answer:

q=1.95*10^{-7}C

Explanation:

According to the free-body diagram of the system, we have:

\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)

So, we can solve for T from (1):

T=\frac{mg}{cos(15^\circ)}(3)

Replacing (3) in (2):

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e

The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)

Finally, we replace (5) in (4) and solving for q:

mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C

5 0
3 years ago
A roller coaster car of mass m= 300 kg is released from rest at the top of a 60 m high hill (position A), and rolls with a negli
Andrews [41]

Answer: The principle of conservation of energy, angular speed and centripetal force

Explanation:

At point A, the car experienced maximum of potential energy

As it moves down the hill, the potential energy decreases while the kinetic energy increases.

The maximum kinetic energy of the car is needed for the attainment of enough centripetal force to help the car move through the loop without falling .

4 0
3 years ago
While at a construction site, you see a crane lift a 1000kg steel beam up to a height of 10m in a time period of 5.0 seconds. A
storchak [24]

Answer:

Car has more power output than crane      

Explanation:

We have given that mass of the crane m = 1000 kg

Height through which crane lift the steel beam h = 10 m

Acceleration due to gravity g=9.8m/sec^2

So work done by crane W=mgh=1000\times 9.8\times 10=98000j

Time period is given as t = 5 sec

We know that power P=\frac{W}{t}=\frac{98000}{5}=19600watt

Now mass of the car = 1000 kg

Initial velocity u = 0 m /sec

Final velocity v = 10 m/sec

We know that work done is equal to the change in kinetic energy

So work done =\frac{1}{2}mv^2-\frac{1}{2}mu^2

=\frac{1}{2}\times 1000\times 10^2-\frac{1}{2}\times 1000\times 0^2=50000j

Time ids given as t = 2 sec

So power P=\frac{W}{t}=\frac{50000}{2}=25000watt

So car has more power output than crane

8 0
3 years ago
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