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3 years ago

14

An electron microscope is usually used for microorganisms that are _____. too small to be seen with the unaided eye too small to

be seen with an optical microscope not possible to observe with any instrument

2 answers:

GaryK [48]3 years ago

5 0

<h3><u>Answer;</u></h3>

<em>too small to be seen with an optical microscope </em>

<h3><u>Explanation;</u></h3>

<em><u>An electron microscope is a type of microscope that is used to observe very tiny specimens whose features can not be observed by other types of microscopes.</u></em> It uses a beam of electrons to generate an image of a given specimen whose features can be clearly observed and studied.
<em><u>Electron microscope has very high resolution and magnification as compared to other optical microscope </u></em>hence can be observed in the study of micro-organisms such as viruses which would be difficult to study their features using optical microscopes.

Olenka [21]3 years ago

4 0

An electron microscope is usually used for microorganisms that are too small to be seen with the unaided eye. You can adjust the size of the microorganism to see it more clearly by adjusting its magnification lens.

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When the distance between two charges is halved, the electrical force between them?

Llana [10]

If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
$F= k \frac{q_1 q_2}{r^2}$
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
$r'= \frac{r}{2}$
the magnitude of the force changes as follows:
$F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k \frac{q_1 q_2}{r^2}=4 F$
so, the force increases by a factor 4.

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3 years ago

During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.146 kg baseball crashing through the pane

Nonamiya [84]

Answer:

Impulse, |J| = 0.6716 kg-m/s

Force, F = 63.35 N

Explanation:

It is given that,

Mass of the baseball, m = 0.146 kg

Initial speed of the ball, u = 15.3 m/s

Final speed of the ball, v = 10.7 m/s

To find,

(a) The magnitude of this impulse.

(b) The magnitude of the average force of the glass on the ball.

Solution,

(a) Impulse of an object is equal to the change in its momentum. It is given by :

$J=m(v-u)$

$J=0.146\ kg(10.7-15.3)\ m/s$

J = -0.6716 kg-m/s

or

|J| = 0.6716 kg-m/s

(b) Another definition of impulse is given by the product of force and time of contact.

t = 0.0106 s

$J=F\times \Delta t$

$F=\dfrac{J}{\Delta t}$

$F=\dfrac{0.6716\ kg-m/s}{0.0106\ s}$

F = 63.35 N

Hence, this is the required solution.

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2 years ago

Thoughts about Genetically Modified Crops

Alex787 [66]

I think that GMOs, also known as genetically modified crops or organisms, can be used in good and bad ways. They can provide crops to survive longer, and produce a bigger portion. However, on the other side of that, it is claimed that it creates a large amount of greenhouse gas emissions, which can turn into climate change.

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2 years ago

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14. This illustration shows the path that a ball follows when it is thrown into the air. The speed of the ball's motion changes

miss Akunina [59]

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2 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

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3 years ago