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ivanzaharov [21]
3 years ago
6

What is the resistance in a circuit that carries a 0.75 A current when powered by a 1.5 V battery? A.) 0.50 Ω B.) 1.1 Ω C.) 2.0

Ω D.) 2.3 Ω
Physics
2 answers:
Margaret [11]3 years ago
8 0

It is 2.0 Ω because R=V/I so 1.5V/.75A = 2Ω

vovangra [49]3 years ago
6 0

Answer: Option (C) is the correct answer.

Explanation:

According to ohm's law, current flowing through a circuit is directly proportional to the voltage over resistance.

Mathematically,        I = \frac{V}{R}

where,         I = current

                   V = voltage

                   R = resistance

Therefore, it is given that current is 0.75 A and voltage is 1.5 V. Thus, calculate the resistance as follows.

                         I = \frac{V}{R}

or,                     R = \frac{V}{I}

                                    = \frac{1.5 V}{0.75 A}

                                    = \frac{15 V \times 10}{75 A}

                                    = 2 ohms

Hence, we can conclude that resistance in the circuit is 2 ohms.

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3 0
3 years ago
child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
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Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
3 years ago
An atom of sodium has a larger mass than an atom of neon. How is this possible when neon has so many more valence electrons?
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