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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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A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i
geniusboy [140]

Answers:

a) t=0.311 s

b) a=86.847 m/s^{2}

c) y=1.736 m

d) V=17.369 m/s

Explanation:

For this situation we will use the following equations:

y=y_{o}+V_{o}t+\frac{1}{2}at^{2} (1)  

V=V_{o} + at (2)  

Where:  

y is the <u>height of the model rocket at a given time</u>

y_{o}=0 is the i<u>nitial height </u>of the model rocket

V_{o}=0 is the<u> initial velocity</u> of the model rocket since it started from rest

V is the <u>velocity of the rocket at a given height and time</u>

t is the <u>time</u> it takes to the model rocket to reach a certain height

a is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

V=\frac{V_{1}+V_{2}}{2} (3)

For this rocket is:

V=\frac{27 m/s}{2}=13.5 m/s (4)

Time is determined by:

t=\frac{y}{V} (5)

t=\frac{4.2 m}{13.5 m/s} (6)

Hence:

t=0.311 s (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

y=\frac{1}{2}at^{2} (8)  

We will use y=4.2 m :

4.2 m=\frac{1}{2}a(0.311)^{2} (9)  

Finding a:

a=86.847 m/s^{2} (10)  

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again y=\frac{1}{2}at^{2} but for t=0.2 s:

y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2} (11)

y=1.736 m (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

V= at (13)  

V= (86.847 m/s^{2})(0.2 s) (14)  

Finally:

V=17.369 m/s (15)  

5 0
3 years ago
How does the angle of launch affect the kinetic energy of a rubber band?​
Lady_Fox [76]

Answer:

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.

3 0
3 years ago
HELP !! Maura is deciding which hose to use to water her outdoor plants. Maura noticed that the water coming out of her garden h
MA_775_DIABLO [31]
THE GREEN HOSE:
Define the (x,y) coordinate at a height of 4 feet up from the ground to match where Majra is holding the green hose.
This means that the equation for the green hose is of the form
y = a(x - h)² + 4          (1)

Because water from the green hose lands on the ground 10 feet from where Majra is standing, therefore
y(10) = -4                    (2)

Because the curve passes through (0,0), therefore
ah² + 4 = 0
ah² = - 4                     (3)

To satisfy (2), obtain
a(10 - h)² + 4 = -4
a(10 - h)² = - 8            (4)

Divide (3) by (4).
h²/(10-h)² = 1/2
2h² = (10 - h)² = 100 - 20h + h²
h² + 20h - 100 = 0             

Solve with the quadratic formula.
x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142
Reject the negative solution.
The vertex is at (4.142, 4).

From (3), obtain
a = -4/4.142² = -0.2332

The equation for the green hose is
y = 0.2332(x - 4.142)² + 4

THE RED HOSE
The red hose has a vertex at (3,7), according to the equation y = -(x-3)² + 7.

A graph of y(x) for both hoses is shown in the attached figure.

Answers:
a. The red hose will throw the water higher. 

b. The equation for the green hose is
     y = -0.2332(x - 4.124)² + 4,
     with the origin at a height of 4 feet above ground level.

c. The domain for the green hose that makes sense is 0 ≤ x ≤ 10 feet.
     The corresponding range is -4 ≤ y ≤ 4 feet.


3 0
3 years ago
For a wavelength of 420 nm, a diffraction grating produces a bright fringe at an angle of 24°. For an unknown wavelength, the sa
Inessa [10]

Answer:

649.8420 nm

Explanation:

We have given \Theta _1=24^{\circ} \ and\ \Theta _2=39^{\circ}

\lambda _1=420\ nm we have to find unknown wavelength that is \lambda _2

The condition for constructive interference is dsin\Theta =n\lambda

Where d is distance between slits \Theta is angle of emergence and n is order of maxima

Using constructive interference equation

dsin\Theta _1=n\lambda _1 ---------eqn 1

dsin\Theta _2=n\lambda _2---------eqn 2

On dividing eqn 1 by eqn 2

\frac{\lambda _1}{\lambda _2}=\frac{sin\Theta _1}{sin\Theta _2}

\frac{420}{\lambda _2}=\frac{sin24^{\circ}}{sin39^{\circ}}

\lambda _2=649.8420\ nm  

7 0
3 years ago
Describe how you determine whether an object is in motion
PolarNik [594]

To determine whether an object is in motion or not, you first
need to specify a reference point, because there's no such
thing as "real" motion, only motion relative to something.

Once you've named the reference point, you have to look at
the object at two different times.  Each time you look at it, you
measure its distance and direction from the reference point. 
If there's any difference in these measurements from one time
to the next, then the object has had average motion during the
period between the two observations.

That's the best you can do ... find average motion during some
period of time.  You can never definitely tell whether or not the
object ever stopped during that time.  But you can sneak up on
it by making the time period between the two observations shorter
and shorter.

6 0
3 years ago
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