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lys-0071 [83]
3 years ago
7

From the following statements about mechanical waves, identify those that are true for transverse mechanical waves only, those t

hat are true for longitudinal mechanical waves only, and those that are true for both types of waves.A. In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy.B. Many wave motions in nature are a combination of longitudinal and transverse motion.C. In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy.D. All of the above
Physics
1 answer:
Dominik [7]3 years ago
5 0

'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy' is true for transverse waves only.

'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy' is true for longitudinal waves only.

'Many wave motions in nature are a combination of longitudinal and transverse motion' is true for both longitudinal and transverse waves.

<u>Explanation:</u>

Longitudinal waves are those where the direction of propagation of particles are parallel to the medium' particles. While transverse waves propagate perpendicular to the medium' particles.

As wave motions are assumed to be of standing waves which comprises of particles moving parallel as well as perpendicular to the medium, most of the wave motions are composed of longitudinal and transverse motion.

So the option stating the medium' particle moves perpendicular to the direction of the energy flow is true for transverse waves. Similarly, the option stating the medium' particle moves parallel to the direction of flow of energy is true for longitudinal waves only.

And the option stating that wave motions comprises of combination of longitudinal and transverse motion is true for both of them.

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2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmos
Citrus2011 [14]

Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh

there, no vertical movement

So, on neglect of height term

P_{1}=P+\dfrac{1}{2}\rho v_{1}^2

Where, P= atmospheric pressure

\rho = air density

v = speed

Put the value in the equation

P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)

P_{1}=2126.0\ lb/ft^2

P_{1}=\dfrac{2126.0}{144}

P_{1}= 14.76\ psi

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{2}=P+\dfrac{1}{2}\rho v_{2}^2

Put the value in the equation

P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)

P_{2}=2246.17\ lb/ft^2

P_{2}=\dfrac{2246.17}{144}

P_{2}= 15.59\ psi

(c).  Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{3}=P+\dfrac{1}{2}\rho v_{3}^2

Put the value in the equation

P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)

P_{3}=2257.93\ lb/ft^2

P_{3}=\dfrac{2257.93}{144}

P_{3}= 15.68\ psi

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

5 0
3 years ago
Jupiter contains a number of small moons that are thought to be captured comets or material ejected by collisions with larger ob
-Dominant- [34]

I believe the evidence for this theory is that:

The orbits surrounding Jupiter are highly elliptical which are off the plane of the ecliptic, and many of these moons are retrograde. This is very unlikely for moons or satellites which are formed during the planetary accretion. Hence comes the theory.

7 0
3 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s. How lo
Arlecino [84]

Answer:

t = 4.08 s

R = 40.8 m

Explanation:

The question is asking us to solve for the time of flight and the range of the rock.

Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:

  • Δx = v_0 t + 1/2at²

We have these known variables:

  • (v_0)_y = 0 m/s
  • a_y = -9.8 m/s²
  • Δx_y = -20 m

And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.

  • -20 = 0t + 1/2(-9.8)t²
  • -20 = 1/2(-9.8)t²
  • -20 = -4.9t²
  • t = 4.08 sec

The time it takes for the rock to reach the ground is 4.08 seconds.

Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.

List out known variables:

  • v_0 = 10 m/s
  • t = 4.08 s
  • a_x = 0 m/s

We are trying to solve for:

  • Δx_x = ?

By using the same equation, we can plug these known values into it and solve for Δx.

  • Δx = 10 * 4.08 + 1/2(0)(4.08)²
  • Δx = 10 * 4.08
  • Δx = 40.8 m

The rock lands 40.8 m from the base of the cliff.

7 0
3 years ago
Suppose a man pushes a crate with a force of 20 N north. What is the magnitude and direction of the reaction force?
Goryan [66]

20 N north because He is pushing it 20 newtons north.

5 0
3 years ago
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