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worty [1.4K]
3 years ago
5

You measure the power delivered by a battery to be 5.83 W when it is connected in series with two equal resistors. How much powe

r will the same battery deliver if the resistors are now connected in parallel across it
Physics
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:

P = 23.32 W

Explanation:

In series  

equivalent Resistance

R(eq)=R+R=2R

In parallel equivalent resistance

R(eq) = R*R/(R+R) =R/2

since.

power

P=V² / R

in series

⇒V = √(P*R)

=√(5.83*2R )

=√(11.66R)

in parallel

P = V² / R(eq)

=(√(11.66R)²) / (R/2)

P=11.66 * R * 2/R

P = 23.32 W

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A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force
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Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

                                                   s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}

     Subsequently,

                                                  F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

                                                    v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}

     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 m \ s^{-1}.

                                                    s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

      worker stops pushing it.

4 0
3 years ago
If only 10 pounds is required to lift a 500-lb block, how much chain must be played out to lift the engine 3.0 inches?
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Answer:

150 inches (12.5 ft)

Explanation:

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The following is the equation one needs to solve:

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10 \,lb\,* \,x\,=\,1500\,lb\,in\\x\,=\,\frac{1500\,lb\,in}{10\,lb} \\\\x\,=150\,in

which can also be given in feet as: 12.5 ft

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The answer is A. Hope this helps you with your work.
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