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Dmitry_Shevchenko [17]

4 years ago

9

Two lightbulbs have powers P1 and P2 when separately connected across an ideal battery with potential difference ΔV. The bulbs a

re then connected in series across the same battery. Determine the total power consumption of the bulbs in this configuration. (Use any variable or symbol stated above as necessary.)

1 answer:

jarptica [38.1K]4 years ago

6 0

Answer:

$P'=\dfrac{P_1P_2}{P_1+P_2}$

Explanation:

Lets take

Resistance of bulb 1 =R₁

Resistance of bulb 2 =R₂

As we know that power P

P= ΔV²/R

Given that voltage difference is same for both bulbs

So

P₁R₁= ΔV² --------1

P₂R₂= ΔV² -----------2

When these resistance are connected in series then equivalent resistance R

R=R₁+ R₂

The new power P'

P'=ΔV²/R

P'R=ΔV² ------3

From equation 1 ,2 and 3

P'(R₁+ R₂) = ΔV²

$P'\left(\dfrac{\Delta V^2}{P_1}+\dfrac{\Delta V^2}{P_2}\right)=\Delta V^2$

$P'=\dfrac{P_1P_2}{P_1+P_2}$

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G = 10 N/kg or 10 m/s2

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Answer:

a) $U_{g} = 40\,J$ , b) $\eta = 70\,\%$ , c) $v = 20\,\frac{m}{s}$

Explanation:

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$U_{g} = (0.2\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (20\,m)$

$U_{g} = 40\,J$

b) The efficiency of the bounce is:

$\eta = \left(\frac{14\,m}{20\,m} \right)\times 100\,\%$

$\eta = 70\,\%$

c) The final speed of Danielle right before reaching the bottom of the hill is determined from the Principle of Energy Conservation:

$K = U_{g}$

$U_{g} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot U_{g}}{m} }$

$v = \sqrt{\frac{2\cdot (40\,J)}{0.2\,kg} }$

$v = 20\,\frac{m}{s}$

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3 years ago

Which of the following is NOT a

Liula [17]

Answer:B

Explanation:

6 0

4 years ago

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Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?

Alina [70]

The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
$F= k_e \frac{ q_1 q_2}{r^2}$
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Substituting these numbers into the equation, we get
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3 years ago

Cannonball a is launched across level ground with speed v₀ at an angle of 45. Cannonball b is launched from the same location wi

postnew [5]

Answer:

Cannonball b spends more time in the air than cannonball a.

Explanation:

Starting with the definition of acceleration, we have that:

$a=\frac{\Delta v}{\Delta t}\\\\\Delta t= \frac{\Delta v}{a}$

Since both cannonballs will stop in their maximum height, their final velocity is zero. And since the acceleration in the y-axis is g, we have:

$\Delta t= -\frac{v_{oy}}{g}$

Now, this time interval is from the moment the cannonballs are launched to the moment of their maximum height, exactly the half of their time in the air. So their flying time t_f is (the minus sign is ignored since we are interested in the magnitudes only):

$t_f=2\frac{v_{oy}}{g}$

Then, we can see that the time the cannonballs spend in the air is proportional to the vertical component of the initial velocity. And we know that:

$v_{oy}=v_o\sin\theta\\\\\implies t_f=2\frac{v_o\sin\theta}{g}$

Finally, since $\sin60\°=\frac{\sqrt{3} }{2}$ and $\sin45\°=\frac{\sqrt{2} }{2}$ , we can conclude that:

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In words, the cannonball b spends more time in the air than cannonball a.

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4 years ago

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