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goblinko [34]
3 years ago
12

A particle is acted on by two torques about the origin: τ→1 has a magnitude of 8 N·m and is directed in the positive direction o

f the x axis, and τ→2 has a magnitude of 8.9 N·m and is directed in the negative direction of the y axis. In unit-vector notation, find d⁢ℓ→/d⁢t, where ℓ→ is the angular momentum of the particle about the origin.
Physics
1 answer:
lyudmila [28]3 years ago
5 0

To give solution to the exercise we must use the concepts of Torque, Vector magnitude and vector direction of the forces.

For the given problem we have to

T_i = 8Nm

T_j = -8.9Nm

In this way the torque acting on the particle as a function of distance and time is,

\tau = \frac{dL}{dt} = 8\hat{i}-8.9\hat{j}

The net torque acting on the particle is

\tau_{net} = \sqrt{T_i^2+T_j^2}

\tau_{net} = \sqrt{(8)^2+(-8.9)^2}

\tau_{net} = 11.967Nm

PART B) The direction of the torque is given by,

tan\theta = \frac{y}{x}

\theta = tan^{-1}\frac{y}{x}

\theta = tan^{-1}(\frac{-8.9}{8})

\theta = -48.04\°

Therefore the torque direction is 48.04° below the x axis.

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If a machine is doing 500 J of work with 75 watts of power, how long did it take to complete its task?
kompoz [17]

Answer:

time = 6.67secs

Explanation:

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3 years ago
b) Assume the rod is 0.60 m long and has a mass of 0.50 kg, and the clay blob has a mass of 0.20 kg and moves at an initial velo
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Answer:

The correct answer is "6.96 rad/s".

Explanation:

The given values are:

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L = 0.6 m

Mass,

m₁ = 0.5 kg

m₂ = 0.2 kg

Initial velocity,

V = 8 m/s

Now,

The final angular velocity will be:

⇒ \omega =\frac{6m_1V}{(4m_1+3m_2)L}

By substituting the values, we get

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4 0
3 years ago
An applied pulling force of 100 N is used to accelerate an object to the right along a rough surface that offers 40 N of frictio
blondinia [14]

Answer:

9.8 m/s^2

Explanation:

First of all, we analyze the vertical direction.

Along this direction, the object is in equilibrium, so its acceleration is zero, and therefore  the net force on it is zero. There are only two forces acting on the object vertically: the weight, W (downward), and the normal force, N (upward), and since the net force must be zero, we can write

W=N

And since N = 60 N, the weight of the object is

W = 60 N

From this, we can also find the mass of the object, using the equation:

W=mg \rightarrow m = \frac{W}{g}=\frac{60}{9.8}=6.1 kg

where g=9.8 m/s^2 is the acceleration of gravity.

Now we analyze the forces along the horizontal direction. We have:

A pulling force of F=100 N forward

A frictional resistance of F_f = 40 N backward

So the equation of motion in this direction is

F-F_f = ma

And solving for a, we find the acceleration of the object:

a=\frac{F-F_f}{m}=\frac{100-40}{6.1}=9.8 m/s^2

in the horizontal direction, forward.

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