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3 years ago

2. -/1 pointsSerPSE10 2.7.P.017.MI.

1 answer:

8 0

Answer: -7.86cm/s^2 don’t forget negative sign!

v1=14cm/s
dx=2.97-(-5)=7.97cm
t=2.85s

dx= v1*t+1/2at^2

I’m not writing the unit’s into the equation until the end, but you always should include them in all your calculations!! Units are very important and helpful to solve problems! Anyways, here we go;

7.97=(14*2.85)+1/2a(2.85)^2

a=[(7.97)-(14*2.85)]*2/(2.85)^2

a=(-63.86)/(8.1225)

a=-7.86cm/s^2 don’t forget the negative sign!

Any questions please feel free to ask. Thanks.

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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

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Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

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$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

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$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

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3 years ago

A large pot is placed on a stove and 1.2 kg of water at 14°C is added to the pot. The temperature of the water is raised evenly

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Answer:

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Explanation:

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Answer:

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Explanation:

Lets calculate

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Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

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$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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