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4 years ago

Microorganisms characterized by the absence of a nucleus are called

Physics

1 answer:

Sidana [21]4 years ago

8 0

<h2>Answer: Prokaryotes </h2>

Let's start by explaining that cells are essential microscopic units that make up the living beings, capable of reproducing independently. In their structure they have a region called cytoplasm with an inner nucleus, which contains the genetic information of the cell.

Now, in the case of prokaryotic cells there is no defined nucleus, therefore it is said that it has no nucleus. This means that the cell's genetic material is simpler than the others and is dispersed in the cytoplasm (in the nucleoid).

Due to having a less complex genetic information, prokaryotes are considered the first living organisms that existed on the planet and are mostly bacteria.

In addition, among its main characteristics is that they can feed themselves using organic matter and sunlight by photosynthesis or inorganic matter (autotrophic feeding) and can also feed on other organisms (heterotrophic feeding).

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The power in an electrical circuit is given by the equation P = I2R, where I is the current flowing through the circuit and R is

kiruha [24]

Answer:

Explanation: Here we have given a direct equation . There for no need to worry .

P = I²×R

P = (12)² ×100

P = 14400W = 14.4 kW

For second one

P = I² ×R

200 = I²×150

I = √200/150

I = 1.15 A

5 0

3 years ago

Please help me with question B.

Step2247 [10]

It accelerates in the y component (bc of gravity) AND the x-component (b/c of the friction force).

5 0

3 years ago

Use dimensional analysis to determine how the linear acceleration a in m/s2 of a particle traveling in a circle depends on some,

astraxan [27]

I have to say, i love this kind of problems.

So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:

$[a]=[\frac{distance}{time^2}]$ .

Now, we got the radius

$[r] = [distance]$

the angular frequency

$[\omega] = \frac{1}{s}$

and the mass

$[m]=[mass]$ .

Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.

The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write

$a \propto r$ .

The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared

$a \propto r \omega^2$ .

We are almost there. If this were any other problem, we would write:

$a = A r \omega^2$

where A its an dimensionless constant. Its common for this constants to appears if we need an conversion factor. If we wanted the acceleration in cm/s^2, for example. Luckily for us, the problem states that there is no dimensionless constant involved, so:

$a = r \omega^2$

5 0

3 years ago

A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di

navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$ $0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$ $t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$ }

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$ $V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).

Best of luck

8 0

3 years ago

How is the world do you do this?

Rainbow [258]

You do the net force by subtracting the sides. The direction of the box is moving forward to the right by 10 N.

4 0

3 years ago