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grin007 [14]
3 years ago
11

What current flow (I) is associated with an input voltage of 5.0V and resistors R1 = 1.5 kiloohms and R2 = 0.5 kiloohms? Calcula

te this value in mA.
Physics
1 answer:
Y_Kistochka [10]3 years ago
5 0

Answer:

current in series is 2.50 mA

current in parallel is 13.51 mA

Explanation:

given data

voltage = 5 V

resistors R1 = 1.5 kilo ohms

resistors R2 = 0.5 kilo ohms

to given data

current flow

solution

current flow in series is express as here

current = voltage / resistor    .................1

put here all value  in equation 1

current = 5 / (1.5 + 0.5)

current = 5 / 2.0

so current = 2.50 mA

and

current flow in parallel is express as

current = voltage / resistor   ....................2

put here all value in equation  2

current = 5 / (1/ (1/1.5 + 1/0.5))

current = 5 / 0.37

so current = 13.31 mA

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Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

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(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet =  327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf  = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

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Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

   = 0.00353 × 128 × (327-40)

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Quantity of heat required to melt the bullet:

Q = mH_f

Q = 0.00353 × 24500

   = 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

                                                                      = 216.17 J

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