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4 years ago

A woodworker has made four

Physics

2 answers:

3241004551 [841]4 years ago

7 0

Answer:

I think the answer is b

Explanation:

yes its b I just did this prob on my notebook

Lilit [14]4 years ago

6 0

I think the answer is b

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in coal-fired power plants, coal is burned to produce ____________________, which is then used to turn a turbine.

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50 POINTS!!!!!!!

miv72 [106K]

Answer:

A. The shorter the distance the greater the force, so now the force would have increased because they haven't moved away from each other.

B. The force would decrease because they are now further apart.

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3 0

3 years ago

The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00

Vera_Pavlovna [14]

Answer:

(a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c). The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$

Explanation:

Given that,

Area = 8.50 cm²

Distance = 3.00 mm

Potential = 6.00 V

Distance without discharge = 8.00 mm

(a). We need to calculate the capacitance

Using formula of capacitance

$C_{1}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{1}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{3.00\times10^{-3}}$

$C_{1}=2.5075\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=2.5075\times10^{-12}\times6.00$

$Q=1.5045\times10^{-11}\ C$

(b). We need to calculate the initial stored energy

Using formula of initial energy

$E_{i}=\dfrac{1}{2}\times CV^2$

$E_{i}=\dfrac{1}{2}\times2.5075\times10^{-12}\times36$

$E_{i}=4.5135\times10^{-11}\ J$

(c). We need to calculate the capacitance

Using formula of capacitance

$C_{2}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{2}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{2\times8.00\times10^{-3}}$

$C_{2}=9.403\times10^{-13}\ F$

We need to calculate the final stored energy

Using formula of initial energy

$E_{f}=\dfrac{1}{2}\times \dfrac{Q^2}{C}$

$E_{f}=\dfrac{1}{2}\times\dfrac{(1.5045\times10^{-11})^2}{9.403\times10^{-13}}$

$E_{f}=12.036\times10^{-11}\ J$

(d). We need to calculate the work done

Using formula of work done

$W=E_{f}-E_{i}$

Put the value in the formula

$W=12.036\times10^{-11}-4.5135\times10^{-11}$

$W=7.5225\times10^{-11}\ J$

Hence, (a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c). The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$

5 0

4 years ago

Please I need help........

LenKa [72]

Answer:

7.46 J/kg/K

Explanation:

The heat absorbed or lost is:

q = mCΔT

where m is the mass, C is the heat capacity, and ΔT is the change in temperature.

Given q = 15.0 J, m = 0.201 kg, and ΔT = 10.0 °C:

15.0 J = (0.201 kg) C (10.0 °C)

C = 7.46 J/kg/°C

Which is the same as 7.46 J/kg/K.

7 0

3 years ago

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