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Cloud [144]
3 years ago
9

A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period At 1.90 s, the tire is stopped and spun in the o

pposite (clockwise) direction also at 2.70 rad/s. Calculate the change in the tire's angular velocity Aw and the tire's average angular acceleration aav. (Indicate the direction with the signs of your answers.)
(a) the change in the tire's angular velocity Aco (in rad/s) rad/s
(b) the tire's average angular acceleration aav (in rad/s2) rad/s
Physics
1 answer:
Naddik [55]3 years ago
7 0

Answer:

a) -5.40 rad/s

b) -2.842 rad/s²

Explanation:

The direction is important in dealing with such questions. Clockwise is considered negative and counterclockwise is considered positive

a) Δω = final angular velocity - initial angular velocity

          = -2.70 rad/s - 2.70 rad/s

          = -5.40 rad/s

b) ∝ = Δω/Δt = (-5.40 rad/s)/1.90s = -2.842 rad/s²

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Missing Details, Most Are Approximations,Simplicity

Explanation:

I just had this question

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3 years ago
"a horn emits a frequency of 1000 HZ. A 14 m/s wind is blowing toward a listener. What is the frequency of the sound heard by th
Semenov [28]

Answer:

 f_L = 1000 Hz

Explanation:

given,

speed of wind, = 14 m/s

frequency of horn,f_s = 1000 Hz

speed of sound,V = 344 m/s

frequency heard by the listener

using Doppler effect

f_L = \dfrac{v+v_L}{v+v_s}f_s

f_L is the frequency of the sound heard by the listener

f_s is the frequency of sound emitted by the listener

V is the speed of sound

v_L is the speed of listener

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considering the frame of reference in which wind is at rest now, both listener and the source will be moving at 14 m/s

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now on solving we will get

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hence, the frequency heard by the listener is equal to  1000 Hz

 

7 0
3 years ago
Can y’all please help me with this 3 part question?
klio [65]

Answer:

Vf = 210 [m/s]

Av = 105 [m/s]

y = 2205 [m]

Explanation:

To solve this problem we must use the following formula of kinematics.

v_{f} =v_{o} +g*t

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (released from the rest)

g = gravity acceleration = 10 [m/s²]

t = time = 21 [s]

Vf = 0 + (10*21)

Vf = 210 [m/s]

Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)

The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)

Av = (210 + 0)/2

Av = 105 [m/s]

To calculate the distance we must use the following equation of kinematics

v_{f} ^{2} =v_{o} ^{2} +2*g*y\\\\(210)^{2} = 0 + (2*10*y)

44100 = 20*y

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Describe the energy of a playground swing at its highest point
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In heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 milrutes, then you have to stop for 3.5 minutes, and f
g100num [7]

Average velocity  between t= 0 and t = 7.5 minutes = 7.4 m / s

<h3>Further explanation </h3>

Regular straight motion is the motion of objects on a straight track that has a fixed speed

Formula used

\large{\boxed{\boxed{\bold{S\:=\:v\:\times\:t}}}

S = distance = m

v = speed = m / s

t = time = seconds

The motion of objects can be expressed in graphical form

This relationship graph can be in the form of a graph S-t, v-t or a-t

From the S-t graph, the average speed can be determined by determining the slope distance of the curve line from the distance to time ratio

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From the statement of questions there are 3 stages of time that occur

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move with v = 12 m / s, t = 1.5 min = 90 s

So the distance traveled =

s = v x t

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stop (v = 0) so that s = 0 and t = 3.5 min = 210 s

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move with v = 15 m / s and t = 2.5 min = 150 s

so the distance traveled

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s = 15 x 150

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If we look at the existing chart (attached)

then we can draw a straight line from the starting point to the end point so we get a slash curve that shows the average speed

v average = distance traveled / time taken

v average = 3330/450

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<h3>Learn more </h3>

The distance of the elevator

brainly.com/question/8729508

velocity position

brainly.com/question/2005478

resultant velocity

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Keywords: average velocity, Motion of objects, s-t graph

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