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morpeh [17]
3 years ago
8

A sound wave traveling eastward through air

Physics
2 answers:
Sholpan [36]3 years ago
6 0

Answer : A sound wave traveling eastward through air causes the air molecules to vibrate east and west

Explanation :

We know that the sound wave is a longitudinal wave. The particles of medium travels in the direction of propagation of the wave. The space where the particles are closely spaced is called compression and the space where the particles are far apart is called rarefaction. Also, the distance between two consecutive compression and rarefaction is called the wavelength.

If the sound wave is travelling in eastward direction, then the air molecules will move in the direction of propagation of wave. They will vibrate either in east or west direction.  

So, the correct option is (1) " vibrate east and west ".    

Wewaii [24]3 years ago
5 0
A sound wave traveling eastward through air<span> causes the air molecules to Vibrate east and west. </span>
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Abigail runs one complete lap (400m) around the track, while Gabi runs a 50 meter dash in a straight line. Which runner had a gr
Ipatiy [6.2K]

Answer:

\boxed {\boxed {\sf Gabi \ had \ greater \ displacement}}

Explanation:

Displacement is the change in the position of an object.

Abigail runs a complete lap around the track, which is 400 meters. Even though she ran 400 meters, she has <u>no displacement.</u> If she starts and ends at the same spot, there is no change in position.

Gabi runs a 50 meter dash in a straight line. Gabi has <u>50 meters of displacement</u>. She runs in a straight line and is 50 meters away from where she began.

While Abigail ran the farther distance, <u>Gabi had the greater displacement.</u>

4 0
2 years ago
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The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec
Elanso [62]
On Earth, the period of a pendulum is given by:
T_{earth}=2\pi  \sqrt{ \frac{L}{g_{earth} }
where L is the length of the pendulum and g_{earth}=9.81~m/s^2 is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }
where g_{mars}=3.71~m/s^2 is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} }  =  \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.
6 0
4 years ago
A 7500 kg truck traveling at 5.0 m/s east collides with a 1500 kg car moving
julia-pushkina [17]

Since the collision is inelastic, both vehicle will move with a common velocity of 2.5 m/s in south east direction after collision.

<h3>Conservation of Linear Momentum</h3>

It states that, the sum of the momentum before collision is equal to the sum of the momentum after collision.

Given that a 7500 kg truck traveling at 5.0 m/s east collides with a 1500 kg car moving at 20 m/s in a direction 30° south of west. After collision, the two vehicles remain tangled together.

The given parameters are;

  • M1 = 7500 kg
  • U1 = 5 m/s
  • M2 = 1500
  • U2 = 20 sin 30 = 10 m/s
  • V = ?

The formula to find the speed will be

M1U1 - M2U2 = (M1 + M2)V

7500 × 5 - 1500 × 10 = (7500 + 1500)V

37500 - 15000 = 9000V

22500 = 9000V

V = 22500/9000

V = 2.5 m/s

Therefore, the common speed will be 2.5 m/s and the direction of the car will be toward the south east.

Learn more about Collision here: brainly.com/question/7694106

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5 0
1 year ago
A 25kg mass is suspended at the end of a horizontal, massless rope that extends from a wall on the left and from the end of a se
patriot [66]
<h2>Answer:20.97g N,32.63g N</h2>

Explanation:

We consider the forces at the knot.

The vertical forces are

T_{2}Sin(50^{0}}) is the vertical component of tension T_{2} at the knot.

-25g is the weight of the mass 25Kg acting downwards.

The horizontal forces are

-T_{1} is the tension in the rope acting left.

T_{2}Cos(50^{0}) is the horizontal component of tension T_{2} acting towards right.

Since the knot has no mass,it is always in equilibrium.

So,the sum of forces acting on it will be zero.

Balancing vertical forces gives,

T_{2}Sin(50^{0}})-25g=0

T_{2}=32.63gN

Balancing horizontal forces gives,

T_{2}Cos(50^{0})-T_{1}=0

T_{1}=32.63g\times Cos(50^{0})=20.97gN

7 0
3 years ago
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Johnson is dragging a bag on a ice surface. He pulls on the strap with a force of 112 N at an angle of 45° to the horizontal to
Nady [450]

Answer with Explanation:

We are given that

Force=F=112 N

\theta=45^{\circ}

Distance,s=84 m

Time, t=3.33 minutes

We have to find the work done by Johnson on the bag and the power generated by Johnson.

Work done, W=Fscos\theta

Using the formula

Work done, W=112\times 84cos45=6652.46 J

Power, P=\frac{W}{t}=\frac{6652.46}{3.33\times 60}=33.3 watt

Using 1 minute=60 s

Hence, the power generated by Johnson=33.3 watt

5 0
3 years ago
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