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NNADVOKAT [17]
2 years ago
15

A race car's velocity increases from +28 m/s to +36 m/s over a 2.0-s time interval. What is the car's

Physics
1 answer:
BigorU [14]2 years ago
3 0

Answer:

4 m/s^{2}

Explanation:

Acceleration, a=\frac {v-u}{t}

Where v and u are the final and initial velocities of the race car respectively, t is the time taken for the race car to attain velocity of 36 m/s.

Substituting 36 m/s for v, 28 m/s for u and 2 s for t then

a=\frac {36 m/s-28 m/s}{2}=4 m/s^{2}

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A 2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 Ω has one edge parallel to a long straight wire. The near edge
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Answer:

current in loops is 52.73 μA

Explanation:

given data

side of square a = b  = 2.40 cm = 0.024 m

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to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance    .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c)    ...............b

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt       ...........c

so from equation a we get here current

current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt

current = ( 4π×10^{-7}×0.024 / 2π(1.20×10^{-2}) × ln (0.024 + 0.012/0.012) × 120

solve it and we get current that is

current = 4 ×10^{-7}× 1.09861 × 120

current = 52.73 ×10^{-6}  A

so here current in loops is 52.73 μA

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