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3 years ago

A race car is traveling around a track under what circumstance would you hear the highest pitch

2 answers:

6 0

the pitch which you hear will be greater than 300 Hz since the sound source is approaching you. But once stopped, the pitch will be 300 Hz exactly. So the pitch must be gradually decreasing from above 300 Hz to 300 Hz during the slowing down process.

ivann1987 [24]3 years ago

6 0

Answer:

As the car is moving quickly toward you.

Explanation:

usa test prep

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A piece of rope is pulled by two people in a tug-of-war. each exerts a 400-n force. what is the tension in the rope?

Leokris [45]

Newtons 3.law: Action = Reaction

If a body exerts a force on a rope of 400 N the rope exerts a force on the body of 400N also. So the tension in the rope is 400N. See pictures below.

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4 years ago

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Compare the properties of sodium chloride and sand

Jet001 [13]

I am 11 year old and I don't known the answer to this question

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3 years ago

Set up a parallel circuit with three resistors. Set up the battery to be 9 V, one resistor at 10 Ohms, the other at 21 Ohms, and

valina [46]

In a parallel circuit, the equivalent resistance is the reciprocal of (the sum of the individual reciprocals).

1/R = 1/10 + 1/21 + 1/13

1/R = 0.225 mhos

R = 4.45 ohms

I = V / R

The total current out of the battery is

I = (9v)/(4.45ohms)

I = 2.02 Amperes

As the total current leaves the battery, it splits into 3 paths, and each resistor gets part of it. The 10ohm resistor gets the most current; the 21ohm resistor gets the least current. After flowing through the resistors, the 3 currents join and add up to 2.02 Amperes again, and the same current returns to the battery.

Each resistor has the same 9v of EMF across it.

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3 years ago

A light spring with force constant 3.80 n/m is compressed by 7.20 cm as it is held between a 0.300-kg block on the left and a 0.

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4 years ago

Physics questions , will give brainliest

creativ13 [48]

The applicable relationship here is

... acceleration = rate of change of velocity

4. The slope of the velocity curve is constant, so the acceleration is constant. That slope is 3 m/s in 5 s, or (3 m/s)/(5 s) = (3/5) m/s² = 0.6 m/s²

7. You're looking for a point on the velocity curve where its slope is -2 m/s². That will be somewhere between t=0 and t=4, because slope is positive for t>4. The only available choice in that region is t = 2 s.

8. At 6.00 m/s² for 3 seconds, velocity will change (3 s)×(6.00 m/s²) = 18.00 m/s. That would get you from an initial speed of 3.44 m/s to 21.44 m/s, so clearly 3 s is almost right, but a little too long for the given change in velocity. The best choice is 2.91 s.

If you want to actually figure it out, the change in velocity is 20.9 -3.44 = 17.46 m/s. Using the relation a = ∆v/∆t, we can rearrange it to ∆t = ∆v/a, or

... ∆t = (17.46 m/s)/(6 m/s²) = 2.91 s

9. This problem combines the determination of acceleration with a units conversion problem. Numbers in the problem are given in kph and seconds, and answers are given in m/s². At some point, you need to convert from km/h to m/s. The multiplier for that is (1000 m/km)/(3600 s/h) = 1/3.6 (m·h)/(km·s).

Your change in speed is -24.6 km/h = (1/3.6)·(-24.6) m/s ≈ -6.8333 m/s. When that change in speed occurs over 3.56 seconds, the acceleration is

... (-6.8333 m/s)/(3.56 s) ≈ -1.919 m/s² ≈ -1.92 m/s²

5. At 10 s, the velocity is about 14 m/s. Looking for grid points the curve goes through, we can use (11, 16) and (9, 12). That is, over the 2-second range from 9 s to 11 s, the velocity increases 4 m/s from 12 m/s to 16 m/s. The acceleration is

... ∆v/∆t = a = (4 m/s)/(2 s) = 2 m/s²

6. ∆v/∆t = a = (28 m/s - 0 m/s)/(4.22 s) ≈ 6.6351 m/s² ≈ 6.64 m/s²

10. No change in velocity means the acceleration is 0 m/s².

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3 years ago