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Scilla [17]
2 years ago
13

Juan is standing on the street. An ambulance moves toward him and then passes by. What best describes the pitch that Juan hears?

The pitch remains constant as the ambulance moves toward and then away from Juan. The pitch increases and decreases in a repeated pattern as the ambulance passes Juan. The pitch drops to a lower pitch once the ambulance passes by Juan. The pitch increases to a higher pitch once the ambulance passes by Juan.
Physics
2 answers:
avanturin [10]2 years ago
7 0

Answer:

C) The pitch drops to a lower pitch once the ambulance passes by Juan.

Explanation:

ozzi2 years ago
5 0

Answer: The correct answer is "the pitch drops to a lower pitch once the ambulance passes by Juan".

Explanation:

Doppler effect is the phenomenon in which there is an increase or decrease in the frequency when there is relative motion between the source and listener.

Pitch depends on the frequency. if the frequency increases then pitch increases.

When the source and the listener are moving towards each other then there is increase in the frequency. When the source and the listener are moving away each other then there is decrease in the frequency.

In the given problem, Juan is standing on the street. An ambulance moves toward him and then passes by the pitch drops to a lower pitch once the ambulance passes by Juan.

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A 1200-ft equal-tangent crest vertical curve is currently designed for 50 mi/h. A civil engineering student contends that 60 mi/
coldgirl [10]

Answer:

8.95ft

Explanation:

In order to develop this problem it is necessary to consider two concepts:

The first is the design of vertical curves through the general equation for the length of a curved vertical crest in terms of algebraic differences in grades. The second is the Design Controls for Crest vertical curves table (I attach a table at the end).

The aforementioned equation is given by:

L = \frac{AS^2}{200(\sqrt{h_1}+\sqrt{h_2})^2}

Where,

L = leght of vertical curve

S = Sight distance

A = Algebraic difference in grades

h_1 =Height of eye above roadway

h_2 =height of object above roadway surface

From the table we know that for design speed of 60 mi/h the S is 570 ft, while the value of the rate of vertival curve K, for design speed of 50mi/h is 84.

Then we can calculate the Algebraic difference in grades through:

A= \frac{L}{K}

A = \frac{1200}{84}

A = 14.285

Applying the equation to find h_1 we have:

L = \frac{AS^2}{200(\sqrt{h_1}+\sqrt{h_2})^2}

1200 = \frac{14.32(570)^2}{200(\sqrt{h_1}+\sqrt{2})^2}

Solving forh_1

h_1 = 8.95ft

Therefore the height of the driver's eye is 8.95ft

7 0
1 year ago
A certain force gives object m1 an acceleration of 12.0 m/s2. The same force gives object m2 an acceleration of 3.30 m/s2. What
zhenek [66]

Answer:

a)   a = 4,552 m / s²,   b)  a = 2,588 m / s²

Explanation:

Newton's second law is

          F = ma

          a = F / m

in this case the force remains constant

indicate us

* for a mass m₁

       a₁ = F/m₁

       a₁ = 12, m/ s²

* for a mass m₂

        a₂= 3.3 m / s²

a) acceleration

       m = m₂-m₁

we substitute

        a = \frac{F}{m_2 - m_1}

        1 / a = \frac{m_2}{F}  -  \frac{m_1}{F}

let's calculate

        \frac{1}{a} = \frac{1}{3.3} - \frac{1}{12}

         \frac{1}{a} = 0.21969

         a = 4,552 m / s²

b)   m = m₂ + m₁

     a = F / (m₂ + m₁)

     \frac{1}{a}  =  \frac{m_2}{F} + \frac{m_1}{F}

we substitute

           \frac{1}{a} = \frac{1}{3.3} + \frac{1}{12}

           a = 2,588 m / s²

3 0
1 year ago
Which are made of matter? Check all that apply. air bicycles sound waves people clothes sunlight trees
LUCKY_DIMON [66]
Trees bikes people clothes trees
5 0
1 year ago
Read 2 more answers
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

b) If the temperatures changes, we can still reuse the ideal gas equation above:

\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}

h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m

3 0
2 years ago
Find the mass of a person walking west at a speed of 0.8 m/s with a momentum of 52.0 kg.m/s west.
KIM [24]

Answer:

mass of the person walking to west is 65 kg.

Given:

Momentum = 52 \frac{kg m}{s}

Speed = 0.8 \frac{m}{s}

To find:

Mass of the person = ?

Formula used:

Momentum is given by,

P = m × v

Where, P = momentum

m = mass

v = speed

Solution:

Momentum is given by,

P = m × v

Where, P = momentum

m = mass

v = speed

Mass = \frac{P}{v}

m = \frac{52}{0.8}

m = 65 kg

Thus, mass of the person walking to west is 65 kg.

5 0
1 year ago
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