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klio [65]
3 years ago
12

A child is in danger of drowning in the Merimac river. The Merimac river has a current of 3.1 km/hr to the east. The child is 0.

6 km from the shore and 2.5 km upstream from the dock. A rescue boat with speed 24.8 km/hr (with respect to the water) sets off from the dock at the optimum angle to reach the child as fast as possible. How far from the dock does the boat reach the child?
Physics
1 answer:
kolezko [41]3 years ago
4 0
In a problem where a child is danger form drowning from a river who has a current of 3.1km/hr to east and the child is 0.6km fro the shore and the upstream is 2,5km from the dock. So base on the question the boat with a speed of 24.8 km/hr is 1.9 km because the child is 0.6 km off the dock so 2.5 minus 0.6
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  • The total resistance in the circuit is 16 Ohms.
  • The total current in the circuit is 0.5 Ampere.
  • The current at R_1 is 0.5 Ampere.
  • The current at R_3 is 0.5 Ampere.
  • The voltage drop atR_1 is 4 volts.
  • The voltage drop at R_2 is 2.5  volts.
  • The voltage drop at R_3 is 1.5 volts.
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Given:

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R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

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R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega

  • The total current in the circuit = I

V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A (Ohm's law)

  • For series combinations, the current in each resistor remains the same.

So, the current in R_1, R_2 \&R_3:

I_1= I_2= I_3=I=0.5 A\\

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The current across  R_1 = I = 0.5 A

V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V

  • The voltage drop across at R_2 =V_2

The current across  R_2 = I = 0.5 A

V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V

  • The voltage drop across at R_3 = V_3

The current across  R_3 = I = 0.5 A

V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V

  • The total power consumed by circuit:

P= V\times I \\\\= 0.5 A\times 8 V = 4 watt

  • Power consumed at R_1:

P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt

  • Power consumed at R_2:

P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt

  • Power consumed at R_3:

P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt

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