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3 months ago

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A single resistor is connected across the terminals of a battery. Which one or more of the following changes in voltage and curr

ent leaves unchanged the electric power dissipated in the resistor? (A) Doubling the voltage and reducing the current by a factor of two. (B) Doubling the voltage and increasing the resistance by a factor of four. (C) Doubling the current and reducing the resistance by a factor of four.

1 answer:

Maru [3.3K]3 months ago

7 0

Answer:

Every option provided is accurate

Explanation:

The electrical power dissipated by a single resistor linked to a battery can be expressed as:

$P=VI = I^2 R=\frac{V^2}{R}$

where

V signifies the voltage

I denotes the current

R represents the resistance

Now, let's evaluate each scenario:

A) When the voltage is doubled (V'=2V) while the current is halved (I'=I/2), the resulting power dissipation turns out to be:

$P'=V'I'=(2V)(\frac{I}{2})=VI=P$ --> the power remains the same

B) When the voltage is increased to double (V'=2V) and the resistance quadruples (R'=4R), the new power dissipation becomes:

$P'=\frac{V'^2}{R'}=\frac{(2V)^2}{4R}=\frac{V^2}{R}$ --> the power is unchanged

C) If the current is doubled (I'=2I) while the resistance diminishes to one-fourth (R'=R/4), the new power dissipation is:

$P'=I'^2 R'=(2I)^2(\frac{R}{4})=I^2 R$ --> the power is unchanged

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Answer:

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2)

$y=7.39\ m$

Explanation:

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When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

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Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

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