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2 years ago

How does an eye doctor use a lens

Physics

2 answers:

Anna007 [38]2 years ago

7 0

The way their suppose to do it

My name is Ann [436]2 years ago

5 0

By looking through it probably

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A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?

ladessa [460]

<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity. weight = mg = (0.59 kg) x (9.80 m/s^2) weight = 5.782 N The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>

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2 years ago

The highest element in the hierarchical breakdown of the wbs is

Makovka662 [10]

Work package. Hope this helps!

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2 years ago

Adhesion is when forces of the molecules are more strongly drawn to each other than the molecules in other substances true or fa

SSSSS [86.1K]

Answer:

False

Explanation:

that is cohesion. adhesion is force between dissimilar molecules of a body

6 0

3 years ago

I need help for the first 3 plz.

madam [21]

1) hypothesis
2) data
3) method

I think these are correct.

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2 years ago

Read 2 more answers

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/

Hunter-Best [27]

Answer:

$n_{T} = 31.68\,rev$

Explanation:

The angular acceleration is:

$\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}$

$\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}$

And the angular deceleration is:

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}$

$\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}$

The total number of revolutions is:

$n_{T} = n_{1} + n_{2}$

$n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}$

$n_{T} = 31.68\,rev$

4 0

3 years ago