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cricket20 [7]
2 years ago
5

Which of the following statements characterizing types of waves are true?

Physics
1 answer:
Tasya [4]2 years ago
7 0

Answer:

a and b.

Explanation:

In general  types of wave

1. Transverse wave  :

    In these waves particle are vibrate perpendicular to motion of waves.

 Ex : Electromagnetic wave , Radio wave .

2. Longitudinal wave :

   In these waves particle are vibrate along the motion of waves.

 Ex : Sound wave

Mechanical wave  :

1 .These are transverse wave or Longitudinal wave or combination of them .

2.These waves required medium for propagation.

3. The particle are vibrate perpendicular to motion of waves.

So the option a and b are correct.

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A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from
Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat (U_{g}), in joules, is equal to the sum of the final translational kinetic energy (K), in joules, and work losses due to air resistance (W_{l}), in joules:

U_{g} = K +W_{l} (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l} (2)

Where:

m - Mass of the cat, in kilograms.

g - Gravitational acceleration, in meters per square second.

h - Initial height of the cat, in meters.

v - Final speed of the cat, in meters per second.

If we know that m = 2.70\,kg, g = 9.807\,\frac{m}{s^{2}}, h = 5.20\,m and W_{l} = 120\,J, then the final speed of the cat is:

v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }

v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }

v \approx 7.586\,\frac{m}{s}

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0
2 years ago
Hanging by a thread. Two metal spheres hang from nylon threads and attract each other when brought close together. (i) What can
elena-14-01-66 [18.8K]

Answer:

Explained

Explanation:

i)Two spheres hanging from nylon threads attract each other because either the two spheres are charged with opposite sign or only one of the spheres is charge so the other would be charge by induction of the charged sphere and hence attract each other.

ii)However, when they are touched the charges will be rearranged among the two sphere such that the two sphere have exact same magnitude and sign of charge and now they will repel each other or the  magnitude of charges on the two spheres become zero and they neither attract or repel each other.

5 0
2 years ago
If an electromagnetic wave has components Ey = E0 sin(kx - ωt) and Bz = B0 sin(kx - ωt), in what direction is it traveling?
fomenos

Answer:

Its traveling in the +x direction

Explanation:

The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.

8 0
2 years ago
What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate
Alex_Xolod [135]

Answer:

Mass will be 4.437 kg

Explanation:

We have given force constant k = 7 N/m

Time period of oscillation T = 5 sec

So angular frequency \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec

We know that angular frequency is given by

\omega =\sqrt{\frac{k}{m}}

1.256 =\sqrt{\frac{7}{m}}

Squaring both side

1.577 =\frac{7}{m}

m = 4.437 kg

6 0
2 years ago
If the intensity level by 15 identical engines in a garage is 100 dB, what is the intensity level generated by each one of these
insens350 [35]

To develop this problem it is necessary to apply the concepts related to Sound Intensity.

By definition the intensity is given by the equation

\beta = 10Log(\frac{I}{I_0})

Where,

I = Intensity of Sound

I_0= Intensity of Reference

At this case we have that 15 engines produces 15 times the reference intensity, that is

I= 15I_0

And the total mutual intensity is 100 dB, so we should

\beta = 100-10*log(\frac{15I_0}{I_0})

\beta = 100-10*log(15)

\beta = 100-11.76

\beta = 88.23dB

Therefore each one of these engines produce D. 88dB.

5 0
2 years ago
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