HELP ASAP!! WILL TRY TO GIVE BRAINLIEST
How are ribosomes different from other complex animal cell organelles?

Answers

Answer 1
Most Organelles Have Membranes, Ribosomes Do Not

Other organelles in the cell, such as the mitochondria and lysosomes, are enclosed by lipid membranes that separate them from other structures in the cell. Ribosomes exist as free structures that float throughout the cytoplasm of the cell. They do not have membranes, which allows them to pick up translational RNA released from the nucleus and grab onto free amino acids in order to produce protein chains.

Ribosomes Consist of Two Units

Ribosomes have a two units. The smaller unit reads the messenger RNA and the larger unit functions to link the amino acids to form the protein chain. When a ribosome is not producing proteins, these units are separated. Most other organelles are larger than ribosomes and a cell can hold a few thousand ribosomes.

Hooking Up with the Endoplasmic Reticulum

Ribosomes can become membrane bound by the endoplasmic reticulum, an organelle that serves to package proteins in order for them to be transferred to other areas of the cell or for transport outside of the cell. The ribosomes become attached to only one side of the endoplasmic reticulum and this region is called the rough endoplasmic reticulum.

Free Floating Protein Production

Free floating ribosomes make proteins that are usually used in the cytoplasm of the cell. Free ribosomes are not different from bound ribosomes. The cell can even change the number of ribosomes needed depending on the protein production needs of the cell.

Related Questions

Cuál es la diferencia entre fricción beneficiosa y perjudicial? Pon 5 ejemplos de cada uno.

Answers

Answer:

Difference is given below.

Explanation:

The main difference between beneficial and harmful friction is that beneficial friction is necessary for performing different activities while on the other hand, harmful effect of friction is destroy different parts of products and machines. examples of beneficial friction are walking, holding things, rubbing hands to produce heat, running etc whereas examples of harmful friction are destruction of sole, slipping, tearing of machine's part, Wet roads and Mudslides etc.

A cyclist travels at 7 m/s and has a mass of 75 kg. Calculate the kinetic energy of the
cyclist.
3,675 )
262.5
1,837.5)
525 )

Answers

Answer:

1837.5 J

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

From the question we have

[tex]k = \frac{1}{2} \times 75 \times {7}^{2} \\ = 37.5 \times 49 \\ = 1837.5[/tex]

We have the final answer as

1837.5 J

Hope this helps you

Two kids are playing tag. The first kid is running at an unknown speed chasing the other. The first kid has a mass of 25 kg and a KE of 450 J.
How fast is the first kid running?
I really need help

Answers

Answer:

1mph

Explanation:

An electric geyser consumes 2.2units of electrical energy per hour of its use. It is designed to work on 220v. What is the resistance of this device? Find the cost of energy consumed if each unit costs Rs 6.00.

Answers

Answer:

Explanation:

The resistance of this device is determined by using the formuia from Ohms Law:

V = IR

where; V = voltage

I = current

R = resistance

From the above equation:

R = V/ I ---- (1)

And

I = P/V    --- (2)

If 1 unit = 1 KWH = 3.6 × 10⁶ J

Then;

2.2 units = (3.6 × 10⁶ × 2.2)

= 7.92 × 10⁶ J

So, Energy = 7.92 × 10⁶ J

Also:

Power = energy/time

Power =  7.92 × 10⁶ J / 1 hour

Power = 7.92 × 10⁶ J / 3600 seconds

Power = 2200 Watts

From equation (2)

I = P/V

I = 2200/220

I = 10 A

Recall that:

Resistance R = V/I

Resistance R = 220/10

Resistance R = 22 ohms

SO, if 1 unit costs = Ts 6.00

Then the cost of 2.2 unit = 2.2 ( Rs  6.00)

= Rs 13.20

An airplane is traveling at an altitude of 15,490 meters. A box of supplies is dropped from its cargo hold. What is the cargo's velocity when it hits the ground? (Show your work and do not forget units)

Answers

Answer :

556.59 m/s.

Explanation:

Given that,

An airplane is traveling at an altitude of 15,490 meters.

A  box of supplies is dropped from its cargo hold. We need to find the velocity of cargo when it hits the ground.

The initial velocity of the box is 0 as it as at rest. Let v is the velocity of cargo when it hits the ground. We can find it using third equation of motion as follows :

[tex]v^2-u^2=2as[/tex]

Put u = 0 and a = g

[tex]v^2=2gs\\\\v=\sqrt{2gs} \\\\v=\sqrt{2\times 10\times 15490 } \\\\v=556.59\ m/s[/tex]

So, when it hits the ground its velocity is 556.59 m/s.

You apply a 325 N force to a heavy crate with a rope that makes a 29.0° angle with the horizontal, if you pull the crate a distance of 5.37 m, how much work was done by you?

Answers

Answer:

W = 1526.43 J

Explanation:

Given that,

Force acting on a heavy crate, F = 325 N

The rope will make 29.0° angle with the horizontal

The distance covered by the crate, d = 5.37 m

We need to find the work done by the crate. The work done by an object is given by :

[tex]W=Fd\cos\theta\\\\=325 \times 5.37\times \cos29\\\\=1526.43\ J[/tex]

So, the required work done is 1526.43 J.

When illuminated with monochromatic light, a double slit produces a pattern that is a combination of single-slit diffraction and double-slit interference. This can be easily seen if the separation between the slits and the size of slits are related by simple fractions. Find the ratio of the width of the slits to the separation between them, if the first minimum of the single slit pattern falls on the fifth maximum of the double slit pattern.

Answers

Answer:

The ratio is  [tex]k:d = 1 : 5[/tex]

Explanation:

From the question we are told that

   The first minimum of the single slit pattern falls on the fifth maximum of the double slit pattern.

Generally the condition for constructive interference for as single slit is  

     [tex]ksin(\theta) = n\lambda[/tex]

Here  k is the width of the slit  and n is the order of the fringe and for single slit n =  1 (cause we are considering the first maxima)

Generally the condition for constructive interference for as double slit is    

        [tex]dsin\theta = m\lambda[/tex]

Here  d is the separation between the  slit  and m is the order of the fringe and for double slit  m  =  5  (cause we are considering the first maxima)

=>     [tex]dsin\theta = 5\lambda[/tex]

So

       [tex]\frac{ksin(\theta)}{dsin(\theta)} = \frac{\lambda}{5\lambda}[/tex]

=>    [tex]\frac{k }{d} = \frac{1}{5}[/tex]

So  

      [tex]k:d = 1 : 5[/tex]

In uniform circular motion, the factor that remains constant is.............

Answers

Answer:

When a body is in uniform circular motion, its speed remains constant but its velocity, angular acceleration, angular velocity changes due to change in its direction.

Ans
2075 Set B Q.No. 90 A water reservoir fank of capacity
is situated at a height of 20 m from the water level. W
be the power of an electric motor to be used to fill the
3 hours? Efficiency of motor is 70%.

Answers

Probably the last one

PLEASE HELP MEEEE
marking brainliest ​

Answers

Answer:

Option C

Explanation:

Centripetal acceleration formula regarding velocity and radius is,

[tex]a_c=\frac{v^2}{r}[/tex]

now we know the centripetal acceleration is 9 and the radius is 16 so we plug these values into our formula,

[tex]a_c=\frac{v^2}{r}\\\\9=\frac{v^2}{16} \\\\144=v^2\\\\\sqrt{144}=v \\\\v=12\ m/s[/tex]

so velocity is 12 m/s

Now for the angular velocity, the formula of centripetal acceleration regarding angular velocity and radius is,

[tex]a_c=rw^2[/tex]

we know the centripetal acceleration is 9 and the radius is 16 so plug these values into the formula,

[tex]a_c=rw^2\\\\9=16w^2\\\\0.5625=w^2\\\\\sqrt{0.5625}=w \\\\0.75\ rad/s=w\\[/tex]

so angular velocity is 0.75 rad/s

A force of 10N is making an angle of 300 with the horizontal. Its horizontal components will be

Answers

Answer: 5N

Explanation:

Horizontal component is the force that is applied as a result of the diagonal application of force.

here in this case,

given diagonal force =10N.

angle= 30°

horizontal component is found using trigonometry..

so,

A vector of magnitude 10 N is provided.

It is at an angle of 30° to the + x axis.

To find:

The components along the x and y axis.

Calculation:

Along x axis:

F(x) = F cos(θ)

=> F(x) = 10 × cos(30)

=> F(x) = 10 (√3/2) = 5√3N

Along y axis:

F(y) = F sin(θ)

=> F(y) = 10 sin(30°)

=> F(y) = 10 × ½

=> F(y) = 5 N.

Additional information:

1. A vector is a quantity that can be defined with both magnitude and direction.

2. Since force is a vector, we are able to break it down into components along chosen axis.

QUICKLY I AM IN THE LESSON NOWWW

So I need you to write a hypothesis about finding and marking stars on "Worldwidetelescope", I dont know how to start a hypothesis and a paragraph about what I did in this project. I know what I did but someone have to help me with the hypothesis and how to start the paragraph and please make a list of everything it should include, Thank you!!

Answers

Answer:

I don't really know much about the whole stars thing, but I've written a lot of paragraphs in science

Explanation:

Remember:

A paragraph should be in CER formation: Claim (hypothesis/discovery), Evidence, and ReasoningYour reasoning should never start with I believe or I think, your reasoning and evidence are 100% fact basedThe actual hypothesis should only be 1-2 sentences and be straight to the point

Include:

for your hypothesis, keywords relating to your subject, I'm assuming  for your topic you should include things like marking, telescope, stars, etc.why you believe your hypothesis is true so use prior knowledge to justify your hypothesis

A ship
a sound
wave
sends
straight
to the sea bed and receves the echo
1.5 seconds later. The speed of sound in
water
How deep is the
at that place?
collin
is 1500 m/s.
Sea
sea​

Answers

Answer:

The sea bed is 1125 m deep.

Explanation:

Speed

It's the rate of change of the distance (d) traveled by an object over time (t).

The speed can be calculated by the formula:

[tex]\displaystyle v=\frac{d}{t}[/tex]

It's known the speed of sound in water is v=1500 m/s. A sound wave is sent to the sea bed and bounces back to the ship 1.5 seconds later.

That is the time the wave takes to travel twice the distance from the ship to the sea bed, thus the required distance uses half of that time or t=1.5/2 = 0.75 seconds.

Solving the equation for d:

d = v.t

d = 1500 m/s * 0.75 s

d = 1125 m

The sea bed is 1125 m deep.

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