In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give your answer in units of Amps. An Amp is 1 Coulomb of charge flowing through a cross-sectional area of the wire per second - that's a lot of charge per second and will warm up a typical wire quite a bit! Most devices have circuits with larger resistors - kLaTeX: \OmegaΩ (103 LaTeX: \OmegaΩ) and MLaTeX: \OmegaΩ (106 LaTeX: \OmegaΩ) are common.

Answers

Answer 1

Answer:

The current pass the [tex]R_2[/tex] is  [tex]I = 0.25 A[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image  

From the question we are told that

    The voltage  is  [tex]V = 3V[/tex]

     The first resistance is  [tex]R_1 = 7 \Omega[/tex]

     The second resistance is  [tex]R_2 = 5 \Omega[/tex]

Since the resistors are connected in series their equivalent resistance is  

       [tex]R_{eq} = R_1 +R_2[/tex]

Substituting values

         [tex]R_{eq} = 7 + 5[/tex]

         [tex]R_{eq} = 12 \Omega[/tex]

Since the resistance are connected in serie the current passing through the circuit  is the same current passing through [tex]R_2[/tex] which is mathematically evaluated as

        [tex]I = \frac{V}{R_{eq}}[/tex]

Substituting values  

      [tex]I = \frac{3}{12}[/tex]

      [tex]I = 0.25 A[/tex]

In This Circuit The Battery Provides 3 V, The Resistance R1 Is 7 , And R2 Is 5 . What Is The Current

Related Questions

PIUDICITIS CONSECulvely and Circle your aliswers. Lilyo
proper significant digits.
53. When you turn on your CD player, the turntable accelerates from zero to 41.8 rad/s in
3.0 s. What is the angular acceleration?
or​

Answers

Answer:

The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Explanation:

Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.

It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :

[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]

So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Calculate potential energy of a 5 kg object sitting on 3 meter ledge

Answers

Pe=5*9.8*3=147 joules

Answer:147 joules

Explanation:

Mass=m=5kg

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

Height=h=3 meter

Potential energy=m x g x h

Potential energy=5 x 9.8 x 3

Potential energy=147 joules

The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)

Answers

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

[tex]v(t)=ate^{-6t}[/tex]   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]

hence, the maximum speed is v_max = ((1/6)e^-1)a

If Jim could drive a Jetson's flying car at a constant speed of 440 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 12.0 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days).

Answers

Answer:

t = 2.94 x 10⁶ years

Explanation:

The equation used in the case of constant speed is:

s = vt

t = s/v

where,

s = distance = 12 light years

s = (12 light years)(9.461 x 10¹² km/light year) =  113.532 x 10¹² km

v = speed = 440 km/hr

t = time passed = ?

Therefore,

t = (113.532 x 10¹² km)/(440 km/hr)

t = 2.58 x 10¹¹ hr

Now, converting it to years:

t = (2.58 x 10¹¹ hr)(1 year/8766 hr)

t = 2.94 x 10⁶ years

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