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11111nata11111 [884]

3 years ago

13

Suppose you are taking a walk one day when you see a tree branch snap at its base and begin to rotate downward with the break ne

ar the trunk of the tree as a pivot point. You estimate the length of the branch to be about 3 meters. Estimate the initial angular acceleration of the branch in rad/s2. Can you estimate the mass of the tree branch from this information? Why or why not?

1 answer:

Naya [18.7K]3 years ago

6 0

Answer:

α = 3.27 rad/s²

No

Explanation:

Given that

L= 3 m

Lets take mass of tree = m kg

The torque to mass m = m g .L

The mass moment of inertia of tree ,I

I = mL² kg.m²

We know that

τ = I α

m g .L = mL² .α

α= g/L

Now by putting the value

$\alpha =\dfrac{ 9.81}{ 3}\ rad/s^2$

α = 3.27 rad/s²

No we can find the mass of the from the above information because angular acceleration is not depend on the mass of tree.

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light of a wavelength 600 nm shines on a soap bubble film. For what soap film thickness will destructive interference occur

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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

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Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

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$K$ - Bulk module, measured in pascals.

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$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

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This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

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$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

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A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0

3 years ago

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Answer:

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Explanation:

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Let Energy supplied is E which is equal to the work done

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Thus, the energy supplied is 2.36 x 10^6 J.

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