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lesantik [10]
2 years ago
6

PLEASE HELP The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the w

ave shown?
A.6 Hz
B.0.3 Hz
C.0.7 Hz
D.3 Hz
Physics
2 answers:
d1i1m1o1n [39]2 years ago
7 0

Answer:

F = 0.3\ Hz

Explanation:

Given

See attachment for the graph

Required

Determine the frequency

Frequency (F) is calculated as:

F = \frac{1}{T}

Where

T = Time to complete a period

From the attachment, the wave complete a cycle or period in 3 seconds..

So:

F = \frac{1}{3s}

F = 0.333\ Hz

F = 0.3\ Hz --- Approximated

Gnesinka [82]2 years ago
4 0

Answer:

0.3Hz

Explanation:

I took the test :)

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Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec
Alex

Answer:

ωM = 1.7 rad/s

vM = 3.4 m/s

ωA = 1.7 rad/s

vA =  1.7  m/s

Explanation:

Calculation of the angular speed (ω)

We use the following formula for circular motion:

ω = θ / t Formula (1)

Where:

ω: angular speed (rad/s)

θ:angle that the particle travels (rad)

t: time interval (s)

Equivalence

1 revolution = 2π rad

Data

θ= 1 revolution = 2π rad

t = 3.7 s

We replace data in the formula (1):

ω = (2π rad) / (3.7s )= 1.7 rad/s

Because the angular speed depends only on the number of turns in a time interval then Mary's angular speed (ωM) is equal to Alex's angular speed (ωA).

ωM = ωA = 1.7 rad/s

Calculation of the  tangential speed (v)

We use the following formula for circular motion:

v = ω*r  Formula (2)

Where:

v : tangential speed (m/s)

ω: angular speed ( rad/s)

r  : radius of the circular path (m)

Data

ωM = ωM = ω= 1.7 rad/s

rM = 2m  : Mary's circular path radius

rA = 1m :  Alex's circular path radius

We apply the formula (2) to calculate v:

vM = ω* rM = (1.7) (2) = 3.4 m/s

vA = ω* rA = (1.7) (1) =  1.7  m/s

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3 years ago
An amusement ride consists of a car moving in a vertical circle on the end of a rigid boom. The radius of the circle is 10 m. Th
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Answer:

3.7 kN (Up)

Explanation:

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What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each
suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

  • ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
  • A = area of plates and
  • d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

  • σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
  • a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

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2 years ago
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