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Law Incorporation
2 days ago
6

A wheel completes 5.6 revolutions in 8 seconds.

Physics
2 answers:
Ostrovityanka [3K]2 days ago
5 0

Answer:

86.15\pi rad/min

Explanation: Angular velocity refers to the number of revolutions made in a specific amount of time.

We need to convert the revolutions into radians and the given time from seconds to minutes,

Given;

1rev=2\pi rad\\therefore\\5.6rev=5.6*2\pi rad\\= 11.2\pi rad

Also,

60s = 1 min

so

8s=\frac{8}{60}min\\=0.13min

We subsequently divide the number of radians by the time in minutes.

\omega =\frac{11.2\pi}{0.13min}\\\omega=86.15\pi rad/min

serg [3.5K]2 days ago
3 0

Answer:

the angular velocity in rad per minute would be 263.89 rad/min

Explanation:

Angular Velocity is how quickly an object travels along a circular path, measured in rev/min, rad/m.

Given that the wheel has

rev = 5.6

and time = 8 sec = 8 /60 = 2/15 minute

The angular velocity in rev/min would be;

ω = number of revolutions/time

ω = 5.6 rev /  (2/15 minute)

ω = 5.6 rev x 15/2 minute

ω   = 42 rev/minute

the angular velocity in rev/minute is 42 rev/minute, and to convert it to rad/minute we calculate;

1 rev = 2π rad/min

42 rev/min = 42 rev/min x 2π rad/min

                   = 263.89 rad/min

Therefore the angular velocity in rad per minute would be

263.89 rads/min

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Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
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p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
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Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
The velocity at the stagnation point is zero.

The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
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18 days ago
Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h
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6 days ago
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The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string. how slowly can the ball go through its top
Keith_Richards [3168]
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1 month ago
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
serg [3504]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

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|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} denotes tension (N)

μ refers to linear density (kg/m)

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|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

Distance a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5×10^{-6}  m.

When tension is doubled:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Distance in the same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With the increased tension, it moves \Delta x = 15.4×10^{-5} m

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1 month ago
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