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asambeis [7]
3 years ago
14

You are riding a 450-kg horse at 14.4 km>h east along a desert road. You have inertia equal to 60.0 kg. A police officer driv

ing past (whom you know and who knows your inertia and the horse’s inertia) measures your speed relative to the police car and calculates your kinetic energy to be 16.32 kJ. What possible speed(s) could the police car have been driving at the instant the officer measured your speed?
Physics
1 answer:
Mandarinka [93]3 years ago
6 0

Answer:

v_{p1} = 19.32 m/s  and   v_{p2}  = 27.32 m/s

Explanation:

The kinetic energy has as formula

        K = ½ m v²

Where m is the mass and v the speed of the body.

If the policeman calculates the kinetic energy, let's clear the speed

       v = √ 2K/m

Let's reduce the units to the SI system

       K = 16.32 KJ (1000 J / kJ) = 16.32 10³ J

Let's calculate

      v = RA (2 16.32 10 3/60)

      v = 23.32 m / s

The rider at 14.4 km/h, reduce

      v_{h} = 14.4 km / h (1000m / 1km) (1h / 3600s) = 4.00 m / s

We already have the relative speed of the two (rider and police car) which is 23.32, we also have the rider speed 4.0 m / s, let's calculate the possible police speeds

These two speeds come from going in the same direction or in opposite directions

        v_{r} = v_{h} ± v_{p}

        v_{p}  = v_{r} - vj

        v_{p1}  = 23.32 - 4

        v_{p1} = 19.32 m/s

        v_{p2}  = v_{r} + vj

        v_{p2} = 23.32 +4

        v_{p2}  = 27.32 m/s

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Two resistors have resistances R(smaller) and R(larger), where R(smaller) &lt; R(larger). When the resistors are connected in se
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Answer:

1.61ohms and 4.39ohms

Explanation:

According to ohm's law which States that the current (I) passing through a metallic conductor at constant temperature is directly proportional to the potential difference (V) across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the current

Rt is the effective resistance

Let the resistances be R and r

When the resistors are connected in series to a 12.0-V battery and the current from the battery is 2.00 A, the equation becomes;

12 = 2(R+r)

Rt = R+r (connection in series)

6 = R+r ...(1)

If the resistors are connected in parallel to the battery and the total current from the battery is 10.2 A, the equation will become;

12 = 10.2(1/R+1/r)

Since 1/Rt = 1/R+1/r (parallel connection)

Rt = R×r/R+r

12 = 10.2(Rr/R+r)

12(R+r) = 10.2Rr ... (2)

Solving equation 1 and 2 simultaneously to get the resistances. From (1), R = 6-r...(3)

Substituting equation 3 into 2 we have;

12{(6-r)+r} = 10.2(6-r)r

12(6-r+r) = 10.2(6r-r²)

72 = 10.2(6r-r²)

36 = 5.1(6r-r²)

36 = 30.6r-5.1r²

5.1r²-30.6r +36 =

r = 30.6±√30.6²-4(5.1)(36)/2(5.1)

r = 30.6±√936.36-734.4/10.2

r = 30.6±√201.96/10.2

r = 30.6±14.2/10.2

r = 44.8/10.2 and r = 16.4/10.2

r = 4.39 and 1.61ohms

Since R+r = 6

R+1.61 = 6

R = 6-1.61

R = 4.39ohms

Therefore the resistances are 1.61ohms and 4.39ohms

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