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Schach [20]
3 years ago
13

________ In many cartoon shows, a character runs off a cliff, realizes his predicament, and lets out a scream. He continues to s

cream as he falls. If the physical situation is portrayed correctly, from the vantage point of an observer at the top of the cliff leaning over the edge, the pitch of the scream as he falls should be A) higher than the original pitch and constant. B) higher than the original pitch and increasing as he falls. C) lower than the original pitch and constant. D) lower than the original pitch and decreasing as he falls. E) It is impossible to predict
Physics
1 answer:
IrinaVladis [17]3 years ago
4 0

Answer:

Here the source is moving away from the observer so frequency will be smaller than the actual frequency and since the speed is increasing so the frequency is decreasing with time so correct answer is

D) lower than the original pitch and decreasing as he falls.

Explanation:

As we know by the Doppler's effect of sound we have

so we will have

f = f_o(\frac{v}{v + v_s})

so here when source moves away from the observer with a some speed then the frequency of the sound observed by the observer is smaller than the actual frequency

Here we know that the speed of the source is increasing with time as the source is falling under gravity

So we can say that the pitch of the sound will decrease with time

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We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when ap
marshall27 [118]

Answer with Explanation:

We are given that

Resistance of solenoid,R=4.3 ohm

Magnetic field,B=3.7\times 10^{-2} T

Current,I=4.6 A

Diameter of wire,d=0.5 mm=0.5\times 10^{-3} m

Radius of wire,r=\frac{d}{2}=\frac{0.5\times 10^{-3}}{2}=0.25\times 10^{-3} m

1mm=10^{-3} m

Radius of solenoid,r'=1 cm=1\times 10^{-2} m

1 cm=10^{-2} m

Resistivity of copper,\rho=1.68\times 10^{-8}\Omega m

We know that

R=\frac{\rho l}{A}

Where A=\pi r^2

Using the formula

4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}

l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m

Number of turns of wire=\frac{l}{2\pi r'}

Number of turns of wire=\frac{50.26}{2\pi(1\times 10^{-2}}=800

Hence, the number of turns of the  solenoid,N=799

Magnetic field in solenoid,B=\mu_0 nI

3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6

n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}

n=6404 turns/m

n=\frac{N}{L}

L=\frac{N}{n}

L=\frac{799}{6404}

L=0.125 m=0.125\times 100=12.5 cm

Length of solenoid=12.5 cm

1m=100 cm

8 0
3 years ago
Which of the following statements CANNOT be supported by Kepler's laws of planetary motion?
horsena [70]

Answer:

B) A planet's speed as it moves around the sun will not be the same in six months.

Explanation:

A planet's speed as it moves around the sun will not be the same in six months, is a  statement that CANNOT be supported by Kepler's laws of planetary motion.

8 0
3 years ago
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What body parts were scientists wanting to image that prompted the development of the CT scanner
zalisa [80]

Answer:

The head

Explanation:

4 0
2 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t
nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

\Phi = BA Cos \theta

Here,

\theta = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

\epsilon= -N \frac{d\Phi }{dt}

\epsilon = -N \frac{(BAcos\theta)}{dt}

\epsilon = -NAcos\theta \frac{dB}{dt}

\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)

\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )

At time t = 5.71s,  Induced emf is,

\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)(  (3.05T/s)-(13.9T/s)(5.71s))

\epsilon = 10.9V

Therefore the magnitude of the induced emf is 10.9V

4 0
3 years ago
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A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q
Kryger [21]
D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS
3 0
3 years ago
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