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3 years ago

Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.

Physics

1 answer:

Strike441 [17]3 years ago

8 0

Answer:

The potential difference between the ends of a wire is 60 volts.

Explanation:

It is given that,

Resistance, R = 5 ohms

Charge, q = 720 C

Time, t = 1 min = 60 s

We know that the charge flowing per unit charge is called current in the circuit. It is given by :

I = 12 A

Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :

V = IR

V = 60 Volts

So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.

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How does the suns energy contribute to the carbon cycle

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Answer:

Photosynthesis

Light energy from the sun stimulates photosynthesis by green plants. One major component of the carbon cycle is the reduction of the carbon (IV) oxide content in the atmosphere by plants. Plants take in Carbon (IV) oxide to use during the process of making their own food. During this process light energy is absorbed by chlorophyll and is used to convert carbon (IV) oxide and water to glucose.

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3 years ago

the crust is the thinnest layer on Earth, which allowed it to have a more constant theory of its composition. how do you think s

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3 years ago

Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?

Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

$x_{a}(t) = at+bt^2$

$x_{b}(t) = ct^2-dt^3$

Constants are:

$a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3$

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both $x_a(t)$ and $x_b(t)$ depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

$at+bt^2 = ct^2-dt^3$

$2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0$

$t_1 = 4 - \sqrt{3} = 2.27$ s, $t_1 = 4 + \sqrt{3} = 5.73$ s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

$v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0$

$t_1 = 1$ s, $t_2 = 4.33$ s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

$a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67$ s

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4 years ago

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