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3 years ago

14

ou are to drive to an interview in another town, at a distance of 300 km on an expressway. The interview is at 11:15 a.m. You pl

an to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 40.0 km/h for 43.0 km. What would be the least speed (in km/h) needed for the rest of the trip to arrive in time for the interview?

1 answer:

Shkiper50 [21]3 years ago

8 0

Answer:

133.62 kmh.

Explanation:

Time provided = 3.25 hours.

Distance to be covered 300 km

Times spent in first 100 km = 1 hour

Time spent in next 43 km

= 43 / 40 = 1.075 hours

Total time spent = 2.075 hours

Total distance covered = 143 km

Distance remaining = 300 - 143

=157 km .

Time remaining = 3.25 - 2.075

= 1.175

Speed required = Distance remaining / time remaining

= 157 / 1.175

= 133.62 kmh.

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From the question we are told that

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$A = \pi r^2$

=> $A = 3.142 * 0.009^2$

=> $A = 2.545 *10^{-4}\ m^2$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{dB}{dt}$

Here

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Here $\mu_o$ is the permeability of free space with value

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So

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2 years ago

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Where: F is the force, k is the Coulomb's constant, $q_{1}$ is the value of the first charge, $q_{2}$ is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

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So that,

F = $q_{1}$ $q_{2}$ x $\frac{k}{r^{2} }$

= 2q x $\frac{q}{3}$ x $\frac{k}{r^{2} }$

= $\frac{2q^{2} }{3}$ x $\frac{k}{r^{2} }$

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The electric force between the given charges would change by $\frac{2}{3}$ .

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