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KonstantinChe [14]
3 years ago
5

Are the objects described here in static equilibrium, dynamic equilibrium, or not in equilibrium at all?

Physics
1 answer:
Brut [27]3 years ago
8 0

Answer:

a) dynamic equilibrium; b) No equilibrium; c) Static equilibrium; d) Dynamic equilibrium; e) Not equilibrium; f) Dynamic equilibrium.

Explanation:

a) In this case, the girder is in dynamic equilibrium, although there is movement it is realized at constant speed therefore there is no acceleration and the sum of forces is equal to zero.

b) There is no equilibrium since the velocity is decreasing it slows the girder, therefore the sum of forces will be equal to the product of mass by acceleration according to newton's second law.

c) In this case, the arms of the person are exercising each 100 lb, in order to keep the barbell stable, this allows the static equilibrium.

d) There is movement but however the jet will move at constant speed without any kind of acceleration, therefore it has a condition of dynamic equilibrium.

e) There is no balance as the rock falls and the acceleration due to gravity causes its speed to increase as it goes down.

f) In this case there is dynamic equilibrium as there was a deceleration movement however this force value given by the deceleration multiplied by the mass is not large enough to be able to move the box, due to the weight of the box plus the friction force between the box and the surface.

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During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive
Katyanochek1 [597]

Answer:

The force is F = 1041.7N

Explanation:

The moment of Inertia I is mathematically evaluated as

               I = MR_A^2

Substituting  1.9kg for M(Mass of the wheel) and \frac{66cm}{2} * \frac{1m}{100cm} = 0.33m for R_A(Radius of wheel)

              I = 1.9 * 0.33^2

                = 0.207kgm^2

The torque on the wheel due to net force is mathematically represented as

                      \tau = FR_B  - F_rR_A

Substituting  135 N for F_r (Force acting on sprocket),\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m for R_B (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

                     \tau = F (0.0435) - 135 (0.33)

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of \alpha  = 3.70 rad/s^2 and this torque can also be represented mathematically as

                   \tau = \alpha I

Now equating the two equation for torque

                                F (0.0435) - 135 (0.33) = \alpha I    

Making F the subject

                     F = \frac{\alpha I + (135*0.33) }{0.0435}

Substituting values

                  F = \frac{(3.70 * 0.207)  + (135*0.33)}{0.0435}

                       = 1041.7N

4 0
3 years ago
An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?
Norma-Jean [14]
Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s
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Your answer is B.

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Answer: all of the above and yes

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