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kari74 [83]
4 years ago
9

A 5.30 m long pole is balanced vertically on its tip. What will be the speed (in meters/second) of the tip of the pole just befo

re it hits the ground? (Assume the lower end of the pole does not slip.)
Physics
1 answer:
suter [353]4 years ago
8 0

Answer:

v = 17.66 m/s

Explanation:

As we know that the lower end of the pole is fixed in the ground and it start rotating about that end

so here we can say that the gravitational potential energy of the pole will convert into rotational kinetic energy of the pole about its one end

so we have

mgL = \frac{1}{2}(\frac{mL^2}{3})\omega^2

so we have

\omega = \sqrt{\frac{6g}{L}}

now we have

\omega = \sqrt{\frac{6(9.81)}{5.30}}

\omega = 3.33 rad/s

now the speed of the other tip of the pole is given as

v = \omega L

v = (3.33)(5.30) = 17.66 m/s

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Read 2 more answers
A rigid rectangular loop, which measures 0.30 m by 0.40 m, carries a current of 5.5 A, as shown in the figure. A uniform externa
slamgirl [31]

Answer:

The magnitude will be "(1.097 \ N/m)\hat{j}". The further explanation is given below.

Explanation:

Trying to determine what is usual for a rectangular loop plane,

⇒  \hat{n}=(Cos35^{\circ})(-\hat{i})+(Sin35^{\circ})(\hat{k})

Magnetic moment is given as:

μ = IA\hat{n}

On putting the values in the above formula, we get

μ = (5.5A)[(0.3m)(0.4m)][(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(-\hat{k})]

  = 0.66 Am^2[(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(\hat{k})]

Now the external vector-shaped torque seems to be:

Magnitude,

\sigma=\hat{\mu}\times\hat{\beta}

On putting the values in the above formula, we get

⇒ \sigma=[(0.06Am^2)(Cos35^{\circ}(-\hat{i})+Sin35^{\circ}(-\hat{k})]\times (2.9T)(-\hat{i})

⇒    = (0.66)(2.9)Cos35^{\circ}(\hat{i}\times \hat{i})+(0.66)(2.9)Sin35^{\circ}(\hat{k}\times \hat{k})

⇒    = 0+(1.97 \ N/m)\hat{j}

⇒    =(1.097 \ N/m)\hat{j}

8 0
3 years ago
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