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2 years ago

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Thandy is looking at two cells under the microscope.One is a human cheek cell and the other is a leaf mesophyll cell from a plan

t. Why does thandy need to use a microscope to look at the cells ?

1 answer:

Vanyuwa [196]2 years ago

7 0

Cell are small and us human can't see them with our own eyes. it is in possible to see cell without a microscope

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Which object will sink in freshwater, which has a density of 1.0 g/cm3?

Pani-rosa [81]

Answer:

Object 2, which has a density of 1.9 g/cm3, since it has more density than freshwater.

6 0

2 years ago

Read 2 more answers

The Earth orbits around the sun because the gravitational force that the sun

kotykmax [81]

<h3>Question -:</h3>

The Earth orbits around the sun because the gravitational force that the sun

exerts on the Earth:

O A. causes Earth's acceleration toward the sun.

O B. is very small because the sun is so far from the Earth.

O c. is smaller than the force the Earth exerts on the sun.

O D. pushes the Earth away from the sun.

<h3>Answer -:</h3>

O A. causes Earth's acceleration toward the sun.

<em>I </em><em>hope </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em><em>have </em><em>a </em><em>nice </em><em>time </em><em>ahead!</em>

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2 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

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2 years ago

A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of

vitfil [10]

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

3 0

2 years ago

The velocity of an object is positive and steadily increasing. Which of the following graphs represents how the acceleration of

stiv31 [10]

If an object's velocity is steadily increasing it means that the acceleration is constant at a certain value.

Choice A shows an acceleration of zero which would only be true if the object was not moving or if its velocity was not changing.

Choice B gives us a graph showing acceleration increasing over time and is therefore incorrect.

Choice C is correct because the acceleration is constant. Steadily increasing tells us that the acceleration is fixed at a certain value.

Choice D is incorrect an represents a constant negative acceleration. This would be the case if the object was steadily decreasing in velocity.

4 0

2 years ago