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aalyn [17]
3 years ago
8

Two cars are on the same straight road. Car A moves east at 63 mph and sounds its horn at 638 Hz. Determine the frequency observ

ed by Car B's driver in each of the following cases (sketch every instance showing positions of the cars and the velocities).
(a) Car B is moving east at 78 mph and is behind car A.
(b) Car B is moving east at 78 mph and is in front of car A.
(c) Car B is moving west at 78 mph and is behind car A.
(d) Car B is moving west at 78 mph and is in front of car A.
Physics
1 answer:
LekaFEV [45]3 years ago
5 0

Answer:

6:20 Two cars are on the same straight road. Car A moves east at 55 mph and sounds its horn at 625 Hz. Rank from highest to lowest the frequency observed by Car B's driver in each of the following cases. (Use the following as necessary: a, b, c, and d. Use only->" or .=" symbols. Do not include any parentheses around the letters or symbols. Let "behind" mean car B is west of car A, and "in front of mean car B is east of car A.) (a) Car B is moving east at 70 mph and is behind car A (b) Car B is moving east at 70 mph and is in front of car A (c) Car B is moving west at 70 mph and is behind car A (d) Car B is moving west at 70 mph and is in front of car A. Need Help? Why does a vibrating guitar string sound louder when placed on the instrument than it would if allowed to vibrate in the air while off the This answer has not been graded yet A soft drink bottle resonates as air is blown across its top. What happens to the resonant frequency as the level of fluid in the bottle decreases? This answer has notl been graded yet An airplane mechanic notices that the sound from a twin-engine aircraft varies rapidly in loudness when both engines are running. What could be causing this variation from loud to soft

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Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia
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Answer:

The acceleration of rocket B is -22.24 m/s²

Explanation:

Two rockets are flying in the same direction and are side by side at the

instant their retrorockets fire

That means they started from the same point at the same time

Rocket A has an initial velocity of  5800 m/s

Rocket B has an initial velocity of 8600 m/s

After a time t both rockets are again side by side, the displacement of

each being zero

That means they are in the same position again → s = 0

The acceleration of rocket A is  -15 m/s²

We need to find the acceleration for rocket B

We can find the time from the information of rocket A by using the

rule → s = u t + \frac{1}{2} a t²

where s is the displacement, u is the initial velocity, a is the

acceleration, and t is the time

→ s = 0 , u = 5800 m/s , a = -15 m/s²

Substitute these values in the rule

→ 0 = 5800 t + \frac{1}{2} (-15) t²

→ 0 = 5800 t - 7.5 t²

Add 7.5 t² for both sides

→ 7.5 t² = 5800 t

Divide both sides by 7.5 t

→ t = 773.3 s

The time for the both rocket to have displacement zero is 773.3 s

Now we can find the acceleration of the rocket B by using the same

rule above

→ u = 8600 m/s , t = 773.3 , s = 0

→ 0 = (8600)(773.3) + \frac{1}{2} a (773.3)²

→ 0 = 6650380 + 298996.445 a

Subtract 6650380 from both sides

→ -6650380 = 298996.445 a

Divide both sides by 298996.445

→ a = -22.24 m/s²

<em>The acceleration of rocket B is -22.24 m/s²</em>

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4 years ago
In static electricity what are the subatomic particles, their charges, their location in the atom
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Electrons - have a negative charge and orbit the nucleus
Protons - have a positive charge and are located in the nucleus
Neutrons - have no charge and are also located in the nucleus
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4 years ago
Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00
lions [1.4K]

Answer:

b.  \Delta KE = 390 eV

Explanation:

As we know that the electric field due to infinite line charge is given as

E =\frac{\lambda}{2\pi \epsilon_0 r}

here we can find potential difference between two points using the relation

\Delta V = \int E.dr

now we have

\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr

now we have

\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})

now plug in all values in it

\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})

\Delta V = 216ln6 = 387 V

now we know by energy conservation

\Delta KE = q\Delta V

\Delta KE = (e)(387V) = 387 eV

3 0
3 years ago
A garden hose with a small puncture is stretched horizontally along the ground. The hose is attached to an open water faucet at
tresset_1 [31]

Answer:

The pressure inside the hose 7000 Pa to the nearest 1000 Pa.

<em>h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}</em>

<em>V_{1} =V_{2}</em>

The pressure at the site of the puncture is <em>P₂ =7161.3 Pa</em>

Explanation:

<em>According to Poiseuille's law, P_{1} - P_{2}  = \frac{128μLQ}{πD^{4} } </em>

<em>Where P_{1} is the pressure at a point 1 before the leak, P_{2} is the pressure at the point of  the leak 2, μ = dynamic viscosity, L = the distance between points 1 and 2, Q = flow rate, D = the diameter of the garden hose. </em>

<em>Also,  from the equation P =ρgh, the equations h₁ = \frac{P₁} {ρg} and h₂ = \frac{P₂} {ρg} can be derived.</em>

Combining Poseuille's law with the above, we get h₁ - ρgh₂ =  \frac{128μLQ}{πD^{4} }

<em>h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}</em>

<em>V =\frac{Q}{A}</em>

Since the hose has a uniform diameter, the nozzle at the end is closed and neither point <em>1 nor 2 lie after the puncture,</em>

<em>V_{1} =V_{2}</em>

The pressure at the site of the puncture <em>P₂ =ρgh₂</em>

<em>P₂ =1kgm⁻³ ×9.81ms⁻¹×0.73m</em>

<em>P₂ =7161.3 Pa</em>

<em />

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3 years ago
One hot day, you find yourself going to the refrigerator for a cold drink. In light of the concept of homeostasis, you're acting
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One hot day, you find yourself going to the refrigerator for a cold drink. In light of the concept of homeostasis, you're acting in terms of the drive-reduction theory.

7 0
3 years ago
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