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Fiesta28 [93]
3 years ago
12

A bicycle tire is spinning counterclockwise at 3.00 rad/s. During a time period Δt = 2.30 s, the tire is stopped and spun in the

opposite (clockwise) direction, also at 3.00 rad/s. Calculate the change in the tire's angular velocity Δω and the tire's averageangular acceleration α. (Indicate the direction with the signs of your answers.)
Physics
1 answer:
emmasim [6.3K]3 years ago
6 0

Answer:

Δω =  -6.00 rad/s

α = -2.61 m/s²

Explanation:

Step 1: Data given

A bicycle tire is spinning counterclockwise at 3.00 rad/s

Δt = 2.30 s

In theopposite (clockwise) direction, also at 3.00 rad/s

Step 2: Calculate the change in the tire's angular velocity Δω

Δω =  ωf -  ωi

ωf = - 3.00 rad/s

 ωi  = 3.00 rad/s

Δω =  ωf -  ωi  = -3.00 - 3.00 = -6.00 rad/s

Step 3: Calculate the tire's average angular acceleration α

α =  Δω /  ΔT

α = -6.00 rad/s /2.30s  

α = -2.61 m/s²

A negative angular acceleration means a decreasing angular velocity

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A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

V_{r} = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

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2 years ago
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3 years ago
The equation for the speed of an electromagnetic wave is
igomit [66]

Answer:

c = 1 / √(ε₀*μ₀)

Explanation:

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What physical feature of a wave is related to the depth of the wave base? What is the difference between the wave base and still
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2 years ago
If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by
fgiga [73]

Answer:

a) v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) v=65-32(2)=1ft/s

Explanation:

From the exercise we got the ball's equation of position:

y=65t-16t^{2}

a) To find the average velocity at the given time we need to use the following formula:

v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

--

y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

--

y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

7 0
3 years ago
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