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geniusboy [140]
3 years ago
14

Which of these is a transverse wave? Help pls

Physics
2 answers:
ELEN [110]3 years ago
5 0

Answer:

<h2><em>Option 1 </em></h2>

Explanation:

1st option is the correct answer......

vodka [1.7K]3 years ago
4 0
The correct answer is option 1.
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You are In-line skates at the top of a small hill. Your potential energy is equal to 1000 J. The last time we checked, your mass
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A) <u>Weight = mass × acceleration (due to gravity)  </u>

= 60×9.8  

= 588 N  

<u>B) Potential energy = mass x gravity x change in height </u>

1,000 = 60.0 x 9.8 x h

h =  1.7 m      

<u>C) Kinetic energyF = potential energyI </u>

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Answer: The answer is B. Higher than.

Explanation:

3 0
2 years ago
Read 2 more answers
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

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(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

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(c) EE = ½ kx²

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EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

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2 years ago
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