1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Andrei [34K]
4 years ago
15

Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m

ass, but has hexagonal sections where material has been removed. The attached masses are released from rest and allowed to fall a height h.Which of the following statements about their angular accelerations is true? a. The angular acceleration of the two flywheels is different but it is impossible to tell which is greater. b. The angular acceleration of flywheel A is greater The angular acceleration of flywheel B is greater. c. Not enough information is provided to determine. d. The angular accelerations of the two flywheels are equal.
Physics
1 answer:
inysia [295]4 years ago
4 0

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

      τ = I α

     T r = I α

     T gR = I α

     Alf = T R / I (1)

     T = α I / R

Now let's use Newton's second law in the mass that descends

     W- T = m a

     a = (m g -T) / m

The two accelerations need related

     a = R α

    α = a / R

    a = (m g - α I / R) / m

    R α = g - α I /m R

    α (R + I / mR) = g

    α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

You might be interested in
In reaching her destination, a backpacker walks with an average velocity of 1.22 m/s, due west. This average velocity results be
Vinvika [58]
Wow! 2.95m/s is a mighty fast pace for a backpacker. Must have one of those Star Wars anti-gravity packs. Also, I would be curious as to why she passed her destination and then walked back. 
<span>Anyway, it goes like this: </span>
<span>Say the time walking east is 't', and the total time is 'T'. </span>
<span>Then 5340 m + .511 t = 1.43 T </span>
<span>(This assumes that velocity is positive in both directions) </span>
<span>Two unknowns in one equation. But you also know that the time spent walking west is </span>
<span>5340m/ 2.95m/s = 1810 sec. </span>
<span>and the total time T = 1810 +t </span>
<span>Substitute this into the first equation, and you can solve for t = 3092 sec. </span>
<span>Then T = 4902 sec. and distance walked east is .511t = 1580m.</span>
8 0
4 years ago
A stationary hockey puck has a mass of 0.27 kg. A hockey player uses her stick to apply a 18 N force over a distance of 0.42 m.
jeka57 [31]
W=Fs
W=18(0.42)
W=7.56

Answer: 7.6 J (first option)
3 0
3 years ago
In a skating stunt known as "crack-the-whip," a number of skaters hold hands and form a straight line. They try to skate so that
4vir4ik [10]

Answer:

The magnitude of the centripetal force that acts on him

Explanation:

Given that,

Mass = 80.0 kg

Distance = 6.10 m

Speed = 6.80 m/s

We need to calculate the magnitude of the centripetal force that acts on him

Using formula of the centripetal force

F_{c}=\dfrac{mv^2}{r}

Where, F = force

m = mass

v = speed

r = distance

Put the value into the formula

F_{c}=\dfrac{80.0\times(6.80)^2}{6.10}

F_{c}=606.4\ N

Hence, The magnitude of the centripetal force that acts on him

7 0
3 years ago
two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i
NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

(t - 1) = 0

\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

t^2 - 3t - 4 = 0

(t - 4)(t + 1) = 0

(t - 4) = 0

\boxed {t = 4 ~ seconds}

The difference in the two ball's time in the air is:

\Delta t = 4 ~ seconds - 1 ~ second

\large {\boxed {\Delta t = 3 ~ seconds} }

<h2>Question B:</h2><h3>First Ball</h3>

v^2 = u^2 - 2gH

v^2 = (-14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

<h3>Second Ball</h3>

v^2 = u^2 - 2gH

v^2 = (14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0
3 years ago
Find the ratio of the radii of a baseball to the Earth, knowing that the radius of a baseball is .09 m, and that the Earth's rad
Xelga [282]
0.09 / 6.37 x 10⁶ = 1.4129 x 10⁻⁸

The radius of the baseball is  1.4129 x 10⁻⁸   the radius of the Earth.
6 0
3 years ago
Other questions:
  • A pronghorn antelope has been observed to run with a top speed of 97 km/h. Suppose an antelope runs 1.5 km with an average speed
    13·1 answer
  • Which of the following shows a car holding a steady speed
    9·2 answers
  • A group of organ systems that work together make up a(n)
    15·2 answers
  • I don’t how to woke this
    10·1 answer
  • What is a time paradox?
    14·2 answers
  • A car starts at 25 m/s and then slows down at 13 m/s in 5 seconds. What is the acceleration of the car? What does the accelerati
    13·1 answer
  • Entropy is the amount of heat a system releases. true or false?​
    12·2 answers
  • 63. Which of the following types of validity is estab-
    13·1 answer
  • What is the relationship between friction and velocity?
    14·2 answers
  • A scientist stated that rock layer C is the
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!