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makkiz [27]
3 years ago
10

An amoeba is a one-celled protist. Amoeba contain all the organelles of a typical eukaryotic animal cell. It is a heterotroph an

d moves through an aquatic environment using cytoplasmic streaming and pseudopods What would you expect to happen FIRST if an amoeba senses prey present in its environment?
Physics
1 answer:
horrorfan [7]3 years ago
4 0

The correct answer is the amoeba will deploy its pseudopods (cytoplasmic extentions) to capture the prey and phagocyte.

The amoeba most known and probably the most representative of the kind. Large (up to 500 microns), common in stagnant waters, extremely voracious as evidenced by multiple digestive vacuoles.

Amoebae are characterized by a deformable cell body emitting changes of shape, the pseudopods, which allow them to crawl on a support or to capture microscopic prey by phagocytosis.

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A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th
zhannawk [14.2K]

Answer: V_{f}=2.96m/s    

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight w of the block has an x-component and y-component:

w_{x}=wsin(\theta)    (1)

w_{y}=wcos(\theta)    (2)

As well as the Normal Force N:

N_{x}=Nsin(\theta)    (3)

N_{y}=Ncos(\theta)    (4)

In addition, we know N=w, then \sum F_{y}=0

In the X-component:

\sum F_{x}=m.a

m.a=w_{x}    (5)

Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

So:

m.g.sin(\theta)=m.a   (7)

a=g.sin(\theta)    (8)

a=5.88m/{s}^{2}    (9)   >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

d=\frac{1}{2}a{t}^{2}   (10)

We already know the value of  d and calculated a, we have to find t:

t=\sqrt{\frac{2d}{a}}   (11)

t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

t=0.50s   (13) >>>This is the time it takes to the block to go from the initial velocity V_{o} to its final velocity V_{f}

If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

V_{f}-V_{o}=a.t   (13)

If V_{o}=0

V_{f}=a.t   (14)

V_{f}=(5.88m/{s}^{2})(0.50s)   (15)

Finally we get the value of the Final Velocity of the block:

V_{f}=2.96m/s    

6 0
3 years ago
What happens to the coefficient of friction when the weight is increased? Why is this?
Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0
4 years ago
Read 2 more answers
During fabrication processes, polymeric materials are generally subjected to which of the following conditions? The ambient temp
geniusboy [140]

Answer:

- Elevated temperatures.

- Elevated pressures.

Explanation:

In the manufacture of plastic products are used:

Raw material (pellets) which are the monomers that promote the chemical reaction.

To these are added the charges, in order to reduce the cost of the final product and improve some properties. These charges can be fiberglass, paper, metal structures.

Additives are also added whose mission is to improve or achieve certain properties, such as reducing friction, reducing chemical degradation, increasing electrical conductivity, coloring the product, and all this happens in the presence of a catalyst that is responsible for initiating and accelerating the chemical reaction process.

There are different methods of production of plastics, one of the most frequent is by injection.

Injection is a process that is carried out on machines similar to extrusion machines, in which the spindle, in addition to rotating, has an axial displacement.

In the injection, once the mold is filled, it is separated from the nozzle of the machine, breaking the feed channel. After a certain time, the piece already cooled is demoulded.

High pressures and temperatures are necessary, but parts of good finish and at high production speeds are obtained

3 0
4 years ago
How can what you learned be used in the real world? (Ex. At the grovery store, future career, etc). Please be specific and use c
vivado [14]
I could use math to help with baking and measuring how much I should use for a cake or stuff like that, they also show us how to add our money together to help us with stuff.
Math, is helpful with many jobs, stuff learned in school can help you strategize and school also makes you think outside of the box sometimes, which can help with jobs. Also if you want to do a career with fashion designing or engineering, you can use math for that too. School also gives you the ability to read.
5 0
3 years ago
A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin
Mashcka [7]

Answer:

r_{cm}=[12.73,12.73]cm

Explanation:

The general equation to calculate the center of mass is:

r_{cm}=1/M*\int\limits {r} \, dm

Any differential of mass can be calculated as:

dm = \lambda*a*d\theta  Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}

So, the differential of mass is:

dm = \frac{2M}{a\pi}*a*d\theta

dm = \frac{2M}{\pi}*d\theta

Now we proceed to calculate X and Y coordinates of the center of mass separately:

X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta

Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta

Solving both integrals, we get:

X_{cm}=2*a/\pi=12.73cm

Y_{cm}=2*a/\pi=12.73cm

Therefore, the position of the center of mass is:

r_{cm}=[12.73,12.73]cm

5 0
3 years ago
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