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3 years ago

10

I am in love guess who?

1 answer:

forsale [732]3 years ago

5 0

Answer: yo mama

Some home sweet Alabama thing?

Explanation:

You might be interested in

A dog in an open field runs 11.0 m east and then 28.0 m in a direction 49.0 ∘ west of north. In what direction must the dog then

Ede4ka [16]

Answer:

$\theta=19.03 SE$

Explanation:

From the question we are told that:

$d_1=11.0m East$

$d_2=28.0m 49 \textdegree west\ north$

$d_3=10m south$

Generally the equation for Resolutions is mathematically given by

$d_x=-11-(-28sin49)$

$d_x=10.132m$

Where

$d_y=-11-(28cos49)$

$d_y=-29.37m$

Generally the equation for Direction is mathematically given by

$\theta=tan^{-1}(\frac{dx}{dy})$

$\theta=tan^{-1}(\frac{10.132}{29.37})$

$\theta=19.03 SE$

7 0

3 years ago

The density of the Earth is greater than the density of an equal volume of water by _____.

defon

Answer:

I think the second option is correct

please mark me as brainliest

5 0

3 years ago

Lenny loves physics and math. In which Energy career pathway would these interests be the most helpful?

vazorg [7]

Answer:

it is D

Explanation:

cause i took the test

3 0

4 years ago

Read 2 more answers

I need help idk if it’s C or D?

alex41 [277]

I think the answer is D.

6 0

3 years ago

A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodie

vivado [14]

Answer:

The length of the rod for the condition on the question to be met is $L = 1.5077 \ m$

Explanation:

The Diagram for this question is gotten from the first uploaded image

From the question we are told that

The mass of the rod is $M = 3.41 \ kg$

The mass of each small bodies is $m = 0.249 \ kg$

The moment of inertia of the three-body system with respect to the described axis is $I = 0.929 \ kg \cdot m^2$

The length of the rod is L

Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

$I = I_r + 2 I_m$

Where $I_r$ is the moment of inertia of the rod about the describe axis which is mathematically represented as

$I_r = \frac{ML^2 }{12}$

And $I_m$ the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as

$I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}$

Thus $2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}$

Hence

$I = M * \frac{L^2}{12} + m * \frac{L^2}{2}$

=> $I = [\frac{M}{12} + \frac{m}{2}] L^2$

substituting vales we have

$0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2$

$L = \sqrt{\frac{0.929}{0.40867} }$

$L = 1.5077 \ m$

6 0

3 years ago