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Ede4ka [16]
3 years ago
11

A loudspeaker in a parked car is producing sound whose frequency is 20 510 Hz. A healthy young person with normal hearing is sta

nding nearby on the sidewalk but cannot hear the sound because the frequency is too high. When the car is moving, however, this person can hear the sound. (a) Is the car moving toward or away from the person
Physics
1 answer:
Vsevolod [243]3 years ago
4 0

Answer:

The car must be moving away from the person.

Explanation:

From Doppler's Effect, we know that when a sound source moves towards a stationary observer, the apparent frequency of that sound increases. While the apparent frequency decreases if the source moves away from the stationary observer.

The audible range of frequencies for a human ear is 20 Hz to 20000 Hz. Therefore, in order for the sound of a loud speaker to be audible for the person, the frequency must decrease below 20000 Hz.

<u>Due to this reason, the car must be moving away from the person.</u>

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An unlabeled hierarchical diagram of various astronomical bodies is shown below. The labels A, B, C, and D can be used to repres
Paha777 [63]
A) The biggest astronomical object is the Universe, which contains billions of galaxies among which there is the Milky Way.
The Milky Way contains thousands of planetary systems, among which the Solar System.
The Solar System contains many <span>planets <span>(but only one star, the Sun)</span>,</span> among which there is Earth.
Therefore you can label:
A = Universe, B = Milky Way, C = Solar system, D = Earth

b) Given what we said before, you could label D also any other planet in the Solar System, therefore you can choose among Mercury, Venus, Mars, Jupiter, Saturn, Uranus, and Neptune.
8 0
3 years ago
A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o
Marrrta [24]

Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

a)

Initial energy

KE_i=\dfrac{1}{2}mu^2

KE_i=\dfrac{1}{2}0.3\times 10^2

KEi= 15 J

Final kinetic energy

KE_f=\dfrac{1}{2}mv^2

KE_f=\dfrac{1}{2}0.3\times 5^2

KEf=3.75 J

The  energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

b)

The minimum value to complete the circular arc

 V=\sqrt{r.g}

Now by putting the values

V=\sqrt{0.8\times 10}

V= 2.82 m/s

So kinetic energy KE

KE=\dfrac{1}{2}mV^2

KE=\dfrac{1}{2}0.3\times 2.82^2

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

                                                                                                    = 22.5 J

Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

4 0
3 years ago
A massless string connects a 10.00 kg mass to a 13.00 kg cart which is resting on a frictionless horizontal surface. The mass ha
ch4aika [34]

The cart's acceleration to the right after the mass is released  is determined as 7.54 m/s².

<h3>Acceleration of the cart</h3>

The acceleration of the cart is determined from the net force acting on the mass-cart system.

Upward force = Downward force

ma = mg

13a = 10(9.8)

13a = 98

a = 98/13

a = 7.54 m/s²

Thus, the cart's acceleration to the right after the mass is released  is determined as 7.54 m/s².

Learn more about acceleration here: brainly.com/question/14344386

#SPJ1

6 0
2 years ago
You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -61.0 units. A) Assuming the x co
kvasek [131]

Answer:

A) magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

B)22.5° (clockwise form -ve X axis)

Explanation:

given vector = V1 = x i - 60 j

magnitude of V1 = 90

x - component can be found out by resultant formula

90^2 = x^2 + (-60)^2

x = 67.08 = 67.1 units (3sf)

FOR THE VECTOR 80 UNITS IN -VE X DIRECTION

The X component is -80-------(1)

The Y component is 0 ---------(2)

<u>For the x- component of new added vector:</u>

(1)----------- x + 67.1 = -80

x = -147.1 = -147.1

<u>For the y- component of new added vector:</u>

<u>(</u>2)---------- y - 61  = 0

y = 61.0 (3sf)

the new added vector is  = -147.1 i + 61 j

magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

direction = arctan (61 / 147.1)

               = 22.5° (clockwise form -ve X axis)

<u />

4 0
3 years ago
A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.
N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2

(c) Moment of inertia :

The net torque acting on it is, \tau=I\alpha, I is the moment of inertia

Also, \tau=Fr

So,

I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2

Hence, this is the required solution.

3 0
3 years ago
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