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melamori03 [73]

3 years ago

9

The driver of a 1000 kg car traveling on a highway at 35 m-s'applies brake to avoid hitting a van in front of him, which had com

e to rest because of trans ahead. After the brakes are applied, a constant frictional force of 8000 N on the car car 35 m- 1,8 1000kg BOOON 52.1 State the WORK- ENERGY THEOREM in words 22 Using ENERGY PRINCIPLES, determine the minimum distance requi for the brakes of the car to be applied to avoid a collision with the van?

1 answer:

puteri [66]3 years ago

6 0

Answer:

Explanation:

kinetic energy will be converted to work of friction

½mv² = Fd

d = mv²/2F = 1000(35²) / (2(8000)) = 76.5625 ≈ 77 m

will be the minimum stopping distance d

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(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
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If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
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$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

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A capacitance C and an inductance L are operated at the same angular frequency.

Levart [38]

A) $\omega = \frac{1}{\sqrt{LC}}$

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$X_L = \omega L$

where L is the inductance.

Since we want the two reactances to be equal, we have

$X_C = X_L$

So we find

$\frac{1}{\omega C}= \omega L\\\omega^2 = \frac{1}{LC}\\\omega = \frac{1}{\sqrt{LC}}$

B) 7449 rad/s

In this case, we have

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Therefore, substituting in the formula for the angular frequency, we find

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C) $39.5 \Omega$

Now we can us the formulas of the reactances written in part A). We have:

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$X_C = \frac{1}{\omega C}=\frac{1}{(7449 rad/s)(3.40\cdot 10^{-6}F)}=39.5 \Omega$

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$X_L = \omega L=(7449 rad/s)(5.30\cdot 10^{-3}H)=39.5 \Omega$

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