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3 years ago

7

A scientist is using an objective lens with 40X magnification on his microscope. If the ocular lens magnifies 10X, what is the t

otal magnification being used to visualize the specimen?

1 answer:

zubka84 [21]3 years ago

8 0

Answer:

400x

Total magnification = 40 × 10= 400x

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Steam is more dangerous than the boiling water because it has more a.Temperature

balandron [24]

Answer:

c) heat

Explanation:

steam has more heat energy than boiling water due to its latent heat of vaporisation.

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3 years ago

The diagram below represents an electromagnetic wave. Please Help<br>

forsale [732]

Answer:

I'm pretty sure it's A

Explanation:

bc the crest is is the point on the wave

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3 years ago

An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min. (a) Calculate the magnitude of the centripe

jonny [76]

To solve this problem it is necessary to apply the concepts related to the relationship between tangential velocity and centripetal velocity, as well as the kinematic equations of angular motion. By definition we know that the direction of centripetal acceleration is perpendicular to the direction of tangential velocity, therefore:

$a_c= \frac{v^2}{r}=r\omega^2$

Where,

V = the linear speed

r = Radius

$\omega =$ Angular speed

The angular speed is given by

$\omega =6500\frac{rev}{min}(\frac{2\pi rad}{1 rev})(\frac{1 min}{60s})$

$\omega = 680.6784 rad/s.$

Replacing at our first equation we have that the centripetal acceleration would be

$a_c = r\omega^2$

$a_c = (0.0750)(680.6784)^2$

$a_c = 34 749.23 m/s^2$

To transform it into multiples of the earth's gravity which is given as $9.8m / s$ the equivalent of 1g.

$a_c =34749.23 \frac{m}{s^2} (1\frac{g}{9.8m/s^2})$

$a_c = 3545.84g$

PART B) Now the linear speed would be subject to:

$v = \omega r$

$v= (680.6784)(0.0750)$

$v=51.05 m/s$

Therefore the linear speed of a point on its edge is 51.05m/s

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3 years ago

what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo

Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are; </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

$The \ moment \ of \ inertia \ of \ uniform \ sphere \ is \ I = \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}$

Which gives;

$\mathbf{I_{Earth}} = \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}$

$\mathrm{The \ rotational \ kinetic \ energy \ is} \ E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}$

$\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}$

Therefore;

$\omega = \dfrac{2 \cdot \pi}{24} = \dfrac{\pi}{24}$

Which gives;

$\mathbf{E_{rotational}} = \dfrac{1}{2} \times 9.7202107 \times 10^{37} \times \left( \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}$

The rotational kinetic energy of the Earth, $E_{rotational}$ = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

3 0

3 years ago

Arlecino [84]

Answer:

3.844 x 10 to the 5th power

Explanation:

4 0

3 years ago