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4 years ago

Can someone help with this question ??

Physics

1 answer:

Marysya12 [62]4 years ago

5 0

Answer:

don't know sorry pls try another web page

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Before a cell enters mitosis, what happens to the genetic information in that cell?

trasher [3.6K]

Before a cell could undergo to the process of mitosis, the cell will have to undergo first to the stage of interphase. Where it will need to pass though three stages where it will prepare before it could undergo to cell division. The cell will most likely pass to G1, S and G2 phase for it to be checked and prepared before it passes through the mitosis.

4 0

3 years ago

A 77.0 kg firefighter climbs a flight of stairs 8.0 m high. how much work is required? 0 incorrect: your answer is incorrect. j

Alex73 [517]

<span>6160 joules

to lift 1 newton 1 metre requires 1 joule
there are 10 newtons in one kilo

so 77(kg) x 8 (metres) x 10 (newtons/kilo) = 6160 joules</span>

5 0

4 years ago

A foot spa machine is an electronic gadget used for soaking,bathing and massaging the feet.

OverLord2011 [107]

Answer:

Just gonna take this at free points and yes you are right but I am confused on what you wanted us to do

4 0

3 years ago

If the length of a building is 2 1 2 times the width and each dimension is increased by 7 ft, then the perimeter is 266 ft. Find

N76 [4]

Answer:

The original dimensions of the building is 95 ft × 38 ft.

Explanation:

Let the original length be 'l' and original width be 'w'.

Given:

Original length (l) = $2\frac{1}{2}\times original\ width$

Original width = 'w'.

So, $l=2\frac{1}{2}w=\frac{5}{2}w$

Now, as per question:

Length and width is increased by 7 ft.

So, new length (l') = $l+7=\frac{5w}{2}+7$

New width (w') = $w+7$

New perimeter (P) = 266 ft

Perimeter is given as:

$P=2(l' +w')\\\\266=2(\frac{5w}{2}+w)\\\\\frac{266}{2}=\frac{5w+2w}{2}\\\\266=7w\\\\w=\frac{266}{7}=38\ ft$

Therefore, original width = 38 ft.

Original length is, $l=\frac{5\times 38}{2}=\frac{190}{2}=95\ ft$

Hence, the original dimensions of the building is 95 ft × 38 ft.

7 0

4 years ago

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

dem82 [27]

Answer:

$r_\alpha=16cm$

Explanation:

The radius of the circumference described by a particle in a cyclotron is given by:

$r=\frac{mv}{qB}(1)$

m is the particle's mass, v is the speed of the particle, q is the particle's charge and B is the magnitude of the magnetic field.

Kinetic energy is defined as:

$K=\frac{mv^2}{2}=\frac{m^2v^2}{2m}\\$

Solving this for mv:

$mv=\sqrt{2mK}(2)$

Replacing (2) in (1):

$r=\frac{\sqrt{2mK}}{qB}$

For protons, we have:

$r_p=\frac{\sqrt{2m_pK}}{eB}(3)$

For alpha particles, we have:

$r_\alpha=\frac{\sqrt{2m_\alpha K}}{(2e)B}(4)$

Dividing (4) in (3):

$\frac{r_\alpha}{r_p}=\frac{\frac{\sqrt{2m_\alpha K}}{(2e)B}}{\frac{\sqrt{2m_p K}}{(e)B}}\\r_\alpha=\frac{r_p}{2}\sqrt{\frac{m_\alpha}{m_p}}\\r_\alpha=\frac{16cm}{2}\sqrt{\frac{6.64*10^{-27}kg}{1.67*10^{-27}kg}}\\r_\alpha=\frac{16cm}{2}(\sqrt{3.98})\\\\r_\alpha=\frac{16cm}{2}(2)\\r_\alpha=16cm$

4 0

4 years ago