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tigry1 [53]
3 years ago
15

A battery with an emf of 4 V and an internal resistance of 0.7 capital omega is connected to a variable resistance R. Find the c

urrent and power delivered by the battery when R is (a) 0, I = 5.714285714 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 5.714285714 OK P = 0 W * [1.25 points] 3 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0 OK (b) 5 capital omega, I = 0.701754386 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.701754386 OK P = 2.462296091 W * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 2.462296091 OK (c) 10 capital omega, and I = 0.3738317757 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.3738317757 OK P = 1.397501965 W * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 1.397501965 OK (d) infinite. I = 0 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0 OK P = W
Physics
1 answer:
Zinaida [17]3 years ago
4 0

Answer:

E = I(R + r)

Making I the subject of the formular by dividing both sides by R + r,

I = E/(R + r)

E = 4V, r = 0.7Ohm, R = 0

I = 4/(0 + 0.7) = 4/0.7

I = 5.174285714A

Explanation:

For a cell of emf E, internal resistance r, connected to an external resistance R, the current flowing through the circuit will be given as:

I = E/(R + r). I is measured in Amperes(A), emf in volts(V), R in Ohms and internal resistance r also in ohms

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Answer:

a) 0.32 m b) -2.4 m c) 1.08 m/s d) -4 m/s

Explanation:

a)

  • As the x and y axes (as chosen) are perpendicular each other, the movements along these axes are independent each other.
  • This means that we can use the kinematic equations for displacements along both axes.
  • In the x direction, as the only initial velocity is in the south direction (-y axis), the skateboarder is at rest, so we can write:

        x =\frac{1}{2}*a*t^{2} (1)

  • In the y-direction, as no acceleration is acting on the skateboarder, we can write  the following displacement equation:

        y = v_{0y} * t  (2)

  • For t = 0.6s, replacing by the givens, we get the position (displacement from the origin) on the x-axis, as follows:

       x =\frac{1}{2}*a*t^{2} =\frac{1}{2} * 1.8 m/s2*(0.6s)^{2}\\ x = 0.32 m

b)

  • From (2) we can get the position on the y-axis (displacement from the origin) as follows:

        y = v_{0y} * t  =  -4 m/s * 0.6 s = -2.4 m

c)

  • In the x- direction, we can find the component of the velocity along this direction, as follows:

        v_{fx} = a*t

  • Replacing by the values, we have:

        v_{fx} = a*t = 1.8 m/s2 * 0.6 s = 1.08 m/s

d)

  • As the skateboarder moves along the y-axis at a constant speed equal to her initial velocity, we  have:

        vfy = voy = -4 m/s

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At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant accele
EleoNora [17]

Answer:

306.8264448 m

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Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

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s_c=ut+\dfrac{1}{2}at^2\\\Rightarrow s_c=\dfrac{1}{2}at^2

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s_t=ut

In order to overtake both distances should be equal

\dfrac{1}{2}at^2=ut\\\Rightarrow \dfrac{1}{2}at=u\\\Rightarrow t=\dfrac{2u}{a}\\\Rightarrow t=\dfrac{2\times 23.5}{3.6}\\\Rightarrow t=13.056\ s

s_c=\dfrac{1}{2}at^2\\\Rightarrow s_c=\dfrac{1}{2}3.6\times 13.056^2\\\Rightarrow s_c=306.8264448\ m

The distance the car has to travel is 306.8264448 m

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 3.6\times 306.8264448+0^2}\\\Rightarrow v=47.0016\ m/s

The speed of the car when it overtakes the truck is 47.0016 m/s

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