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2 years ago

Can you plz make me brainliast?

Physics

1 answer:

bekas [8.4K]2 years ago

8 0

Answer:

I dont know what to answer so im just gonna say ok

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What is the Genetic Modification Process?<br><br> Brainlyist for the best answer

Dovator [93]

Answer:

Production of GMOs is a multistage process which can be summarized as follows:

1. identification of the gene interest;

2. isolation of the gene of interest;

3. amplifying the gene to produce many copies;

4. associating the gene with an appropriate promoter and poly A sequence and insertion into plasmids;

5. multiplying the plasmid in bacteria and recovering the cloned construct for injection;

6. transference of the construct into the recipient tissue, usually fertilized eggs;

7. integration of gene into recipient genome;

8. expression of gene in recipient genome; and

9. inheritance of gene through further generations.

6 0

2 years ago

Read 2 more answers

A British newspaper was interested in whether groups of people who differed in social standing were treated differently when the

fiasKO [112]

Answer:

That there should be no distinction between a person of certain social standing than another form a different one. They all should be treated the same.

Explanation:

Unfortunately, this experiment showed the reality of today´s society: that eventhough racism, social distiction, bullying, is seen as wrong, it is still happening.

Social standing is very important in the bussiness world and service facilities. People still treat better some community or type of person because they think they are "better" for having something or its believes, it´s in their culture.

In this example, the priest had better service because society thinks that people church-related are better because of their values, type of life, etc.

However, the type of life some people have won´t have to matter when a service is provided. All arethe same.

6 0

3 years ago

Please help with Physics Circuits!

Zigmanuir [339]

1) Let's start by calculating the equivalent resistance of the three resistors in parallel, $R_2, R_3, R_4$ :
$\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}$
From which we find
$R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
$R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega$
So, the correct answer is D)

2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$

And these are connected in series with a resistor of $10 \Omega$ , so the equivalent resistance of the circuit is
$R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega$

And by using Ohm's law we find the current in the circuit:
$I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A$
So, the correct answer is C).

3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
$R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega$

And by using Ohm's law we find the current flowing in the circuit:
$I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A$

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
$V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V$
So, the correct answer is D).

4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}$
From which we find
$R_{23} = 7.55 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is:
$R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega$

The current in the circuit is given by Ohm's law
$I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A$

Now we can compare the voltage drops across the resistors. Resistor 1:
$V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V$
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
$V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V$
Resistor 4:
$V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V$

So, the greatest voltage drop is on resistor 1, so the correct answer is D).

5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
$P_3 = I_3^2 R_3 = \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W$
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.

5 0

3 years ago

Read 2 more answers

Given three vectors A = 24i + 33j, B = 55i - 12j and C = 2i + 43j (a) Find the magnitude of each vector. (b) Write an expression

slega [8]

Answer:

(a) , . and .

(b) $\vec A - \vec C=22 \hat i -10 \hat j$ .

Explanation:

Given that,

and

$\vec {C}=2 \hat i +43 \hat j$

(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.

So, the magnitude of the is

$|\vec A|=\sqrt {24^2+ 33^2}$

$\Rightarrow |\vec A|=\sqrt {1665}$

The magnitude of the is

$|\vec B|=\sqrt {55^2+ (-12)^2}$

$\Rightarrow |\vec B|=\sqrt {3169}$

And, the magnitude of the is

$|\vec C|=\sqrt {2^2+ 43^2}$

$\Rightarrow |\vec C|=\sqrt {1853}$

(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

$\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)$

$\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j$

$\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j$

$\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)$

$\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j$

$\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)$

The magnitude of is

$|\vec A - \vec B|=\sqrt {(-31)^2+55^2}$

$\Rightarrow |\vec A - \vec B|=\sqrt {3986}$

$\Rightarrow |\vec A - \vec B|=63.13$

Now, if a vector $\vec V= -\alpha \hat i +\beta \hat j$ in 3rd quadrant having direction $\theta$ with respect to $\hat i$ direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector $\vec A - \vec C$ , $\alpha=31$ and $\beta=45$ .

$\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)$

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(d) Vector diagrams for $\vec A +\vec B$ and $\vec A - \vec B$ has been shown

in the figure(b) and figure(c) recpectively.

Vector $\vec A - \vec B$ is in 3rd quadrant as calculated in part (c).

While Vector $\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)$

$\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j$ , which is in 1st quadrant as both the components are position has been shown in figure(b).

6 0

3 years ago

The path that thermal energy follows is from ___________ to ___________

Helga [31]

Warmer to colder objects

5 0

3 years ago