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Ne4ueva [31]
3 years ago
11

Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that reduces

the magnitude of a negative velocity? of a positive velocity?
Physics
1 answer:
ladessa [460]3 years ago
8 0

Answer:

Positive sign for negative velocity and minus sing for positive velocity

Explanation:

In the case of the negative velocity, the sign of the acceleration that reduces its magnitude is the positive sign, since being in the opposite direction  to the movement indicates a deceleration or braking. In the case of the positive velocity, the sign of the acceleration that reduces its magnitude is the negative sign, since being in the  opposite direction to the movement indicates a deceleration or braking. We observe that there will always be a reduction in the magnitude of the velocity if the acceleration goes in the opposite direction.

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(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current
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(a) 1.11 \Omega

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

r=\frac{V}{I}

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

r=\frac{20 V}{18 A}=1.11\Omega

(b) 360 W

The power generated by the battery is given by the equation

P=VI

where

V = 20 V is the voltage of the battery

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Substituting into the formula,

P=(20 V)(18 A)=360 W

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The energy dissipated by the internal resistance is given by

E=Pt

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t = 1 s is the time

Solving the equation, we find

E=(360 W)(1 s)=360 J

(d) 1.65 A

The battery is now connected to a R=11 \Omega resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega

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I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A

(e) 29.9 W

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P=I^2 R

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I = 1.65 A is the current

R=11 \Omega is the resistance

Solving the equation, we find

P=(1.65 A)^2(11 \Omega)=29.9 W

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V

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