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2 years ago

Complete the passage to differentiate between longitudinal and transverse waves.

Physics

2 answers:

koban [17]2 years ago

8 0

Answer:

parallel, and perpendicular

Explanation:

astra-53 [7]2 years ago

3 0

Complete sentence:

A longitudinal wave is a type of wave that transfers energy parallel to the direction of wave motion. A transverse wave, on the other hand, is a type of wave that transfers energy perpendicular to the direction of wave motion.

Explanation

the main difference between longitudinal and transverse wave is the direction of their oscillation/vibration. In a longitudinal wave, the oscillation of the wave occurs in a direction parallel to the direction of propagation of the wave, while in a transverse wave, the oscillation occurs in a plane perpendicular to the direction of propagation. An example of transverse waves are the electromagnetic waves, which consists of an electric field and a magnetic field that oscillates perpendicular to the direction of the wave, while an example of longitudinal waves are the sound waves, whose particles vibrate in a direction parallel to the direction of motion of the wave, producing regions of more density of particles (called compressions) and regions of less density of particles (called rarefactions).

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CAN SOMEONE HELP ME PLEASE ASAP

Lady_Fox [76]

The third choice is correct

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3 years ago

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It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

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2 years ago

In a car with a mass of 4,000 KG is accelerating at a rate of 2m/s2 and hits a tree what force does it have

Ilia_Sergeevich [38]

According to Newton's second law of motion, the acceleration of a body is directly proportional to the force acting on the body and inversely proportional to its mass. The formula for this law is

F=ma

=4000kg * 2m/s 2 =8000N

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3 years ago

One kind of slingshot consists of a pocket that holds a pebble and is whirled on a circle of radius r. The pebble is released fr

Debora [2.8K]

Answer:

θ=142.9°

Explanation:

d=1 *r

angle ϕ= 37.1°

the line connecting pebble and target should be tangent to a circle so

cos(180-ϕ-θ)= $\frac{r}{d}$ = $\frac{1}{1}$

∴ θ=180-ϕ- $cos^{-1} (\frac{1}{1} )$

θ= 180-37.1-0

θ=142.9°

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3 years ago

A floating ice block is pushed through a displacement along a straight embankment by rushing water, which exerts a force on the

lions [1.4K]

The question is incomplete. Here is the complete question.

A floating ice block is pushed through a displacement vector d = (15m)i - (12m)j along a straight embankment by rushing water, which exerts a force vector F = (210N)i - (150N)j on the block. How much work does the force do on the block during displacement?

Answer: W = 4950J

Explanation: <u>Work</u> (W), in physics, is done when a force acts on an object that has a displacement form a place to another:

W = F · d

As the formula shows, Work is a scalar product, i.e, it results in a number, so, Work only has magnitude.

Force and displacement for the ice block are in 2 dimensions, then work will be:

W = (210)i - (150)j · (15)i - (12)j

W = (210*15) + (150*12)

W = 3150 + 1800

W = 4950J

During the displacement, the ice block has a work of 4950J

8 0

2 years ago