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AVprozaik [17]
3 years ago
7

Calculate the total resistance in the circuit​

Physics
2 answers:
vazorg [7]3 years ago
4 0

Answer:

The total resistance of a series circuit is equal to the sum of individual resistances. Voltage applied to a series circuit is equal to the sum of the individual voltage drops. The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor.

SOVA2 [1]3 years ago
4 0

If you know the total current and the voltage across the whole circuit, you can find the total resistance using Ohm's Law: R = V / I. For example, a parallel circuit has a voltage of 9 volts and total current of 3 amps. The total resistance RT = 9 volts / 3 amps = 3 Ω.
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a sprinter with a mass of 80kg accelerates uniformly from 0 m/s to 9 m/s in 3 s. a.)what is the runners acceleration? b.) what i
Katyanochek1 [597]
Part A: 
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
             a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time

Substituting,
            a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
<em>ANSWER: 3 m/s²</em>

Part B: 
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
               F = m x a
where F is force,
m is mass, and 
a is acceleration

Substituting,
              F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N

<em>ANSWER: 240 N </em>

Part C: 
The distance that the sprinter travel is calculated through the equation,
         d = V₀t + 0.5at²

Substituting,
            d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
             d = 13.5 m

<em>ANSWER: d = 13.5 m</em>

3 0
3 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
4 years ago
The band gab of semi conductor?
Lerok [7]

Answer:

Properties of semiconductors are determined by the energy gap between valence and conduction bands. To understand, what is semiconductor, we have to define these terms. In solid-state physics, the energy gap or the band gap is an energy range between valence band and conduction band where electron states are forbidden.

8 0
3 years ago
How could you dissolve more solid solute in a saturated solution in a liquid solvent
tankabanditka [31]
If you saturated the solid it will turn into liquid and soon become an air
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Water is contained in a closed, rigid 0.2 m 3 tank at an initial pressure of 5 bar and a quality of 50%. Heat transfer occurs un
elena55 [62]

Answer:

Final mass=0.89kg

Final pressure=5.6bar

Explanation:

To find mass,m=v/v1

But v1=vf + x(vg-vf)

Vf= 0.001093m^3/kg

Vg= 0.3748m^3/kg

V1= 0.001093+0.5(0.3748-0.001093)

V1= 0.225m^3/kg

M= 0.20/0.225 =0.89kg

Final pressure will be:

V/V1= P/P1

Cross multiply

VP1=V1P

P1= 0.225×5/0.2

P1=:5.6 bar

7 0
4 years ago
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