12 days ago

To what potential should you charge a 2.0 μF capacitor to store 1.0 J of energy?

Physics

1 answer:

Maru [2.3K]12 days ago

6 0

E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V

To store 1.0 J of energy, you need to charge it to 1000 volts.

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A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t

inna [2205]

Answer:

The answer to the specified question will be " $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$ ".

Explanation:

Referring to the question,

$\sum F_{x}$

⇒ $TCos \theta-F_{s}=0$

⇒ $T_{m}Cos \theta =F_{s}$ ...(equation 1)

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⇒ $TSin \theta+F_{N}=m_{g}$

⇒ $M_{g}-TSin \theta=F_{N}$ ...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒ $T_{m} Cos \theta = \mu_{s}F_{N}$

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⇒ $T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)$

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4 0

16 days ago

A student with a mass of 66.0 kg climbs a staircase in 44.0 s. If the distance between the base and the top of the staircase is

serg [2593]

$power = 205.8 \: watt \\ solution \\ mass = 66 \: kg \\ time = 44 \: sec \\ distance = 14 \\ now \\ power = \frac{w}{t} \\ \: \: \: \: \: \: \: \: = \frac{f \times d}{t} \\ \: \: \: \: \: = \frac{m \times g \times d}{t} \\ \: \: \: \: \: \: = \frac{66 \times 9.8 \times 14}{44} \\ \: \: \: \: = \frac{9055.2}{44} \\ \: \: \: \: \: = 205.8 \: watt \\ hope \: it \: helps$