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Bezzdna [24]
3 years ago
15

List 3 quantities of waves.

Physics
2 answers:
VARVARA [1.3K]3 years ago
8 0
If you mean like electromagnetic waves then, Mico waves, UV rays, and infrared waves
Ne4ueva [31]3 years ago
5 0
Waves can be described in terms of their speed, amplitude, and wavelength.
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A ball thrown straight up into the air is found to be moving at 6.79 m/s after falling 1.87 m below its release point. Find the
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Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

v^{2}=u^{2}+2as and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

u=\sqrt {v^{2}-2gs}

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

u=\sqrt {6.79^{2}-(2\times 9.81\times 1.87)}=3.068338313 m/s\approx 3.07 m/s

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3 years ago
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3 years ago
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guajiro [1.7K]

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3 0
3 years ago
When an object's final velocity is less than its initial velocity, however, it has ________________ acceleration.
algol [13]

Answer:

The body has negative acceleration PR a deceleration.

Explanation:

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HAVE A GREAT DAY.

4 0
3 years ago
A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc
nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

m_sv_s + m_rv_r = 0

where m_s = 72.8, m_r = 265 are the mass of the swimmer and raft, respectively. v_s = 5.21 m/s, v_r are the velocities of the swimmer and the raft after the run, respectively. We can solve for v_r

265v_r + 72.8*5.21 = 0

v_b = -72.8*5.21/265 = -1.43 m/s

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

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3 years ago
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