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Alex777 [14]
4 years ago
10

A 15. μF capacitor is connected to a 50. V battery and becomes fully charged. The battery is removed and a slab of dielectric th

at completely fills the space between the plates is inserted. If the dielectric has a dielectric constant of 5.0, what is the voltage across the capacitor's plates after the slab is inserted?
Engineering
1 answer:
Stolb23 [73]4 years ago
3 0

Answer:

voltage across the capacitor's is 75 μF

Explanation:

given data

capacitor = 15 μF

voltage = 50V

dielectric constant k = 5

to find out

voltage across the capacitor's

solution

we will find here voltage across the capacitor's  by this formula

voltage across the capacitor's  = kC_{o}

here k is 5 and C_{o} = 15

put these value we get

voltage across the capacitor's  = kC_{o}

voltage across the capacitor's  = 5 ( 15 )  = 75 μF

so voltage across the capacitor's is 75 μF

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Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

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3 0
3 years ago
Steam enters a heavily insulated throttling valve at 11 MPa, 600°C and exits at 5.5 MPa. Determine the final temperature of the
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Answer: the final temperature of the steam 581.5 °C

Explanation:

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3 years ago
an electric circuit includes a voltage source and two resistances (50 and 75) in parallel. determine the voltage source required
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Answer:

The voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

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Total resistance, R = (50 x 75)/(125)

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V = 120 V

Therefore, the voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

This voltage is also the same for 50 ohms resistance but the current will be 2.4 A.

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