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irina [24]
3 years ago
11

A wave has a distance between its peaks of 0.4 meters at a speed of 10 meters per second. What is the wave's frequency, in hertz

?
Physics
1 answer:
REY [17]3 years ago
8 0

Answer: 25Hz

Explanation:

Given that:

Wavelength (λ) = 0.4m (the distance between the peaks of a wave is known as wavelength)

Frequency F = ?

Speed of wave V = 10 m/s

The wave frequency is the number of cycles which the wave complete in one second. It is measured in hertz, and represented by the symbol F

So, apply the formula V = F λ

10m/s = F x 0.4m

F = 10m/s / 0.4m

F = 25Hz

Thus, the frequency of the wave is 25 Hertz.

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High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of th
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Lifetime = 4.928 x 10^-32 s  

Explanation:

(1 / v2 – 1 / c2) x2 = T2

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3 years ago
Give reason Pascal is a derived unit​
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Answer:

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Explanation:

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4 0
3 years ago
A ball is thrown straight up with a velocity of 50 m/s.(use g = -10 m/s^2) a) What is the velocity in m/s after 2 seconds?
Andrei [34K]

Answer:

(a) 30 m/sec

(b) -50 m/sec

Explanation:

We have given initial velocity of ball u = 50 m/sec

Acceleration due to gravity g=-10m/sec^2

(a) Time t = 2 sec

Now according to first equation of v = u-gt

So v=50-10×2=30 m/sec

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Now according to first equation of motion

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3 0
3 years ago
A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

  = 207 × 1.342 + 1/2 × 896.65 × 1.342²

  = 277.725 + 807.417

  = 1085.14 revolutions ≅ 1085 revolutions

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