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Novosadov [1.4K]
2 years ago
6

Which phrase best describes wave motion

Physics
2 answers:
zavuch27 [327]2 years ago
5 0

Answer:

propagation

Explanation:

The simplest types of wave motion are vibrations of elastic media, such as air, crystalline solids, or stretched strings.

White raven [17]2 years ago
4 0

Answer:

I would describe wave motion as “propagating oscillations” as a general phrase.

For something like water waves, I would say “water waves are oscillations of the surface of a body of water that propagate in a given direction.”

For electromagnetic waves I would say “An electromagnetic wave is an oscillating disturbance of the electric and magnetic fields that propagates through space.”

Explanation:

PLZ MARK BRAINLIEST

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Sonja [21]

Answer:

c

Explanation:i don’t know

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3 years ago
4. What is the momentum of a 70 kg object traveling at 20 m/s?
Soloha48 [4]

Answer:

1400 units of momentum.

Explanation:

Using the formula p=mv. We can get the momentum using 70*20 =1400 units of momentum

6 0
3 years ago
A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i
laiz [17]

(a) Let <em>x</em> be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work <em>W</em> done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to <em>x</em> is

<em>W</em> = - (1/2 <em>kx</em> ² - 1/2 <em>k</em> (0.0400 m)²)

(note that <em>x</em> > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to <em>x</em>, so

<em>W</em> = ∆<em>K</em> = 0 - 1/2 <em>m</em> (0.835 m/s)²

Solve for <em>x</em> :

- (1/2 (160 N/m) <em>x</em> ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==>   <em>x</em> ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

<em>W</em> = - 1/2 <em>k</em> (0.0400 m)²

If <em>v</em> is the glider's maximum speed, then by the work-energy theorem,

<em>W</em> = ∆<em>K</em> = 1/2 <em>m</em> (0.835 m/s)² - 1/2 <em>mv</em> ²

Solve for <em>v</em> :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) <em>v</em> ²

==>   <em>v</em> ≈ 1.43 m/s

(c) The angular frequency of the glider's oscillation is

√(<em>k</em>/<em>m</em>) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

3 0
2 years ago
Similarity between mass and charge
IrinaK [193]
I think there is only one.
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</span>Both are conserved<span>.</span>

5 0
3 years ago
Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be
miv72 [106K]

consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s         (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

h = (0) (3.0) + (0.5) (9.8) (3.0)²

h = 44.1 m

4 0
3 years ago
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