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3 years ago

Calculate the capacitance of a system that stores 9.4 x 10-10 C of charge at

Physics

2 answers:

pantera1 [17]3 years ago

7 0

The capacitance of the system is $1.9\cdot 10^{-11}F$

Explanation:

The capacitance of a capacitor is given by the following equation:

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

where

C is the capacitance

Q is the charge stored in the capacitor

V is the potential difference across the capacitor

In this problem, we have:

$Q=9.4\cdot 10^{-10}C$ is the charge stored

$V=50.0 V$ is the potential difference across it

Substituting into the equation, we find the capacitance:

$C=\frac{9.4\cdot 10^{-10}}{50.0}=1.9\cdot 10^{-11}F$

Learn more about capacitors:

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mart [117]3 years ago

5 0

Answer:

The capacitance of the system is 1.9*10-11

Explanation:

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3 years ago

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

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1 year ago

A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b

Marrrta [24]

Answer:

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Explanation:

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0.0012 * 0.25 m = 0.0003 m

Hooke's law:

F = x * k

Where x is the change in length

F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)

The compressive load will generate a stress of

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3 years ago

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IgorC [24]

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Explanation:

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3 years ago

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