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3 years ago

PLEASE HELP

2 answers:

4 0

Don't over think it!

For example, if you do 5 laps in 5 minutes then you do one lap per minute. If you use that example and you apply it to your problem you'll notice that the answer is simply C. 1km/min.

Best wishes!

Grace [21]3 years ago

4 0

C. 1km/min because the car covers a distance of 5km over a period of 5 minutes, so the car travels 1 kilometer each minute.

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Javier wants to improve his flexibility and increase his personal discipline. Which community resource should he select?

Dovator [93]

Karate class, kicking high and ducking fast makes you much more flexible and you must pay respects to those ou fight with and be intune with yourself to do it right, in other words karate is the answer.

7 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

3 years ago

Alex787 [66]

Explanation:

Volume of 10 coins = 100ml - 75ml = 25ml

Volume of 1 coin = 25ml / 10 = 2.5ml

The average volume of each coin is 2.5ml.

8 0

3 years ago

Read 2 more answers

How is the path of the water particles near the water's surface different from the path near the ocean floor

swat32

Answer: Cold Ocean Currents: It is the layer of the ocean water which moves due to the stress of blowing the wind and this motion is thus called as Ekman Transport. Deep Water Currents: They constitute about 90% of the ocean water. They move around the ocean basin due to variations in the density and gravity.

7 0

3 years ago

A series LRC circuit consists of a 12.0-mH inductor, a 15.0-µF capacitor, a resistor, and a 110-V (rms) ac voltage source. If th

finlep [7]

Answer:

61.85 ohm

Explanation:

L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm

Let ω0 be the resonant frequency.

$\omega _{0}=\frac{1}{\sqrt{LC}}$

$\omega _{0}=\frac{1}{\sqrt{12\times 10^{-3}\times 15\times 10^{-6}}}$

ω0 = 2357 rad/s

ω = 2 x 2357 = 4714 rad/s

XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm

Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm

Impedance, Z = $\sqrt{R^{2}+\left ( XL - Xc \right )^{2}}$

Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm

Thus, the impedance at double the resonant frequency is 61.85 ohm.

7 0

3 years ago