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omeli [17]
3 years ago
6

A police car is parked by the side of a highway and its siren emits a sound measuring 750 Hz. As you approach the police car fro

m a distance, you hear a change in the sound. The change is in the sound's
Physics
2 answers:
Nikitich [7]3 years ago
8 0

As you approach the police car with blaring siren, you hear a change in pitch and frequency. The apparent pitch changes as a result of the Doppler effect. Frequency determines pitch; frequency and wavelength are inversely proportional. You wouldn't "hear" either one. The result is pitch.

D) pitch and frequency.

tangare [24]3 years ago
5 0

Answer:

The correct answer would be frequency, intensity, and amplitude.

It can be explained with the help of Doppler effects which states that the frequency or wavelength of sound changes (increases or decreases) as source and observer move towards or away from each other.

Due to this effect, the frequency and intensity of siren increase as we move towards the siren.

In addition, amplitude also increases as we move towards the source of the sound and decreases as the observer moves away from the source.

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Given the height of rod,length of shadow of tree and length of shadow of the rod, estimate the height of the tree? Given:height
diamong [38]

Answer:

1000 cm.

Explanation:

To obtain the estimated tree height :

(Height of rod / length of rod shadow) = (height of tree / length of tree shadow)

Substituting values into the formula :

(150cm / 120 cm) = (height of tree / 800 cm)

Using cross multiplication :

Height of tree * 120 = 150 * 800

Height of tree = (150 * 800) / 120

Height of tree = 120,000 / 120

Height of tree = 1000

Hence, estimate height of tree = 1000 cm

5 0
3 years ago
A 3.10-mm-long, 430 kgkg steel beam extends horizontally from the point where it has been bolted to the framework of a new build
Alex17521 [72]

Answer:

Explanation:

Given that,

The length of the beam L = 3.10m

The mass of the steam beam m_1 = 430kg

The mass of worker m_2 = 69.0kg

The distance from  the fixed point to centre of gravity of beam = \frac{L}{2}

and our length of beam is 3.10m

so the distance from  the fixed point to centre of gravity of beam is

\frac{3.10}{2}=1.55m

Then the net torque is

=-W_sL'-W_wL\\\\=-(W_sL'+W_wL)

W_s is the weight of steel rod

=430\times9.8=4214N

W_w is the weight of the worker

=69\times9.8\\\\=676.2N

Torque can now be calculated

-(4214\times1.55+676.2\times3.9)Nm\\\\-(6531.7+2637.18)Nm\\\\-(9168.88)Nm

≅ 9169Nm

<h3>Therefore,the magnitude of the torque is 9169Nm</h3>

4 0
4 years ago
A car accelerates from rest. It reaches a velocity of 25m/s in 10 seconds. What was the cars acceleration?
mixer [17]

Answer:

We are Given:

initial velocity (u) = 0m/s

final velocity (v) = 25 m/s

time (t) = 10 seconds

acceleration (a)  = a m/s/s

From the first equation of motion, we know that:

v = u + at

solving for a, we get:

a = (v-u) / t

now, plugging the given values in this equation

a = (25 - 0) / 10

a = 25 / 10

a = 2.5 m/s/s

Therefore, the acceleration of the car is 2.5 m/s/s

3 0
3 years ago
A sound wave of frequency 440 Hz is played by a musician. What is the
TiliK225 [7]

Answer:

343/440

Explanation:

Recall that v=d/t

Now, this is the same thing.

Frequency is 1/T and wavelength is the distance travelled in one period.

So Vs=f*λ

(the greek letter is used as the symbol of wavelength; it's arbitrary)

7 0
3 years ago
A bar-magnet with magnetic moment 2.5 Am^2 is placed in a homogeneous magnetic field (of 0.1 T that is directed along the z-axis
Angelina_Jolie [31]

Answer:

1)

Force on bar magnet  = 0

Torque on bar magnet = 0

2)

Force on bar magnet  = 0

Torque on bar magnet = 0.177 Nm

3)

Force on bar magnet  = 0

Torque on bar magnet = 0.25 Nm

Explanation:

Part 1)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is inclined along z axis along magnetic field

then we will have

\tau = MBsin0 = 0

Part 2)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 45 degree with z axis then we will have

\tau = MBsin45

\tau = (2.5)(0.1)sin45

\tau = 0.177 Nm

Part 3)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 90 degree with z axis then we will have

\tau = MBsin90

\tau = (2.5)(0.1)sin90

\tau = 0.25 Nm

8 0
4 years ago
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