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tankabanditka [31]
2 years ago
6

What is the magnitude of the torque that the axle must apply to prevent the disk from rotating?

Physics
1 answer:
mihalych1998 [28]2 years ago
7 0

The required torque at the axle, is given by the difference between the

moments of the applied forces.

The torque required is <u>19.62 N·m counterclockwise</u>

Reasons:

The given parameters are;

Mass of the disk, m = 5.0 kg

Location of the axle = Half the radius of the disk

Diameter of the disk, D = 40 cm = 0.4 m

Applied mass, 0.1 m from the axle = 15 kg

Applied mass, 0.3 m from the axle = 10 kg

Required:

Magnitude of torque at the axle that prevent the disk from rotating

Solution:

Torque needed = Clockwise moment - Counterclockwise moment

Clockwise moment = (10 kg × 0.3 m + 5 kg × 0.1 m) × 9.81 m/s² = 34.335 N·m

Counterclockwise moment = 15 kg × 0.1 m  × 9.81 m/s² = 14.715 N·m

τ + Counterclockwise moment = Clockwise moment

τ + 14.715 N·m = 34.335 N·m

Torque required, τ = 34.335 N·m - 14.715 N·m = 19.62 N·m

Torque required, τ = <u>19.62 N·m counterclockwise</u>

Learn more here:

brainly.com/question/19044661

brainly.com/question/19247046

<em>The probable question drawing obtained from a similar question online is attached</em>

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an airplane flies at a speed of 100 m/s and starts to accelerate constantly at a rate of 50 m/s2. how fast is the plane flying a
mart [117]

Answer:

331.7m/s

Explanation:

Given parameters:

Initial velocity  = 100m/s

Acceleration  = 50m/s²

Distance  = 1km   = 1000m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we have to apply the right motion equation shown below;

     v²  = u²  + 2aS

 v is the final velocity

 u is the initial velocity

 a is the acceleration

 S is the distance

 Now insert the parameters and solve;

     v² = 100² + (2 x 50 x 1000)

     v² = 110000

     v = √110000  = 331.7m/s

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3 years ago
A good sign that cardiovascular exercise is occurring is that you _____.
grigory [225]
Are breathing hard. This is because cardiovascular exercise makes the heart beat faster which in turn creates a need for more air.
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2 years ago
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2 years ago
Two equally charged, 2.098 g spheres are placed with 3.338 cm between their centers. When released, each begins to accelerate at
photoshop1234 [79]

Answer:

2.64\times 10^{-7} C

Explanation:

There are two spheres name 1 and 2 and they posses the same charge, which is +q.

And they have equal mass which is 2.098 g.

The distance between these two spheres is, r=3.338 cm.

And the acceleration of each sphere is, a=269.429 m/s^{2}.

Now the coulumbian force experience by 1 sphere due to 2 sphere,

F_{21} =\frac{q^{2} }{4\pi\epsilon_{0} r^{2}  }.

And also the newton force will occur due to this force,

F_{21}=ma.

Now equate the above two values of force will get,

\frac{q^{2} }{4\pi\epsilon_{0} r^{2}  } =ma

Further solve this,

q^{2}=ma4\pi  \epsilon_{0} r^{2}.

Substitute all the known variables in above equation,

q^{2}=(2.098\times 10^{-3} )(269.429)(4(3.14))(8.85\times 10^{-12})(3.338\times 10^{-2}).

q=2.64\times 10^{-7} C.

6 0
3 years ago
MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle
labwork [276]

Answer:

(a)  y_{max}=0.423m

(b)  \alpha =64.3^{o}

Explanation:

Given data

v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle

Solution

For Part (a)

As the velocity component in direction of y is given by:

v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s

The maximum displacement is given by:

v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m

For Part (b)

To reach y=46cm =0.46m apply:

0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha  =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}

5 0
3 years ago
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