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Komok [63]
3 years ago
10

A car accelerates at a rate of 3 m/s^2. If its original speed is 8 m/s, how many seconds will it take the car to reach a final s

peed of 25 m/s?
Physics
1 answer:
kkurt [141]3 years ago
4 0

Answer:

\Large \boxed{\mathrm{5.67 \ seconds }}

Explanation:

\displaystyle \mathrm{acceleration \ = \ \frac{final \ velocity - initial \ velocity }{elapsed \ time}}

\displaystyle A \ = \ \frac{V_f - V_i }{t}

\displaystyle 3 \ = \ \frac{25 - 8 }{t}

\displaystyle 3 \ = \ \frac{17 }{t}

\displaystyle t \ = \ \frac{17 }{3} \approx 5.67

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A complete circuit contains two parallel connected devices and a generator for providing the electromotive force. The resistance
irga5000 [103]
We are asked to solve and determine the magnitude of the current flowing through the first device. In order for us to have a better understanding of the problem, we can refer to the attached picture which contains electric circuit diagram. Since it the problem we are already given with an electromotive source or the voltage supply and since the two resistance is in parallel, it would clearly mean that the voltage drop in each resistance is just the same. The resistance 1 uses the 40 volts at the same time the resistance 2 uses 40 volts also. Solving further for the current, we can apply Ohm's law which V = IR where "V" represents the voltage, the "I" represents the current and "R" represents the resistance.

Such as the solution in obtaining current is shown below:
I = V / R, substitute values we have it
I = 40 volts / 1208 ohms
I = 0.0331 Amperes

Therefore, the current flowing in the first device is 0.033 Amperes or 33 milliAmperes.

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3 years ago
A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter
yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

  • Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

        I_{1} = \frac{V}{R_{int} +r_{L} }

  • where Rint = r and RL = r
  • Replacing these values in I₁, we have:

       I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)

  • When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

       I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r}  (2)

  • We can find the relationship between I₂, and I₁, dividing both sides, as follows:

        \frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}

  • The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.  
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1. A metal ball has a mass of 20 g and a volume of 6 cm3.Find its density
vesna_86 [32]

Answer:

density of the ball is 3.33 g/cc

Explanation:

As we know that the density is the ratio of mass and volume

here we know that

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\rho = \frac{m}{V}

\rho = \frac{20}{6} g/cm^3

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