All categories

3 years ago

Chromosomes that carry the same set of genes is called a ________ Chromosome.

Physics

1 answer:

Leokris [45]3 years ago

3 0

It is called a homologous chromosome meaning it carries the same gene

You might be interested in

The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an em

Vilka [71]

Answer:

$\rho=995.50\ kg.m^{-3}$

$\bar w=9765.887\ N.m^{-3}$

$s=0.9955$

Explanation:

Given:

volume of liquid content in the can, $v_l=0.355\ L=3.55\times 10^{-4}\ L$

mass of filled can, $m_f=0.369\ kg$

weight of empty can, $w_c=0.153\ N$

So, mass of the empty can:

$m_c=\frac{w_c}{g}$

$m_c=\frac{0.153}{9.81}$

$m_c=0.015596\ kg$

Hence the mass of liquid(soda):

$m_l=m_f-m_c$

$m_l=0.369-0.015596$

$m_l=0.3534\ kg$

Therefore the density of liquid soda:

$\rho=\frac{m_l}{v_l}$ (as density is given as mass per unit volume of the substance)

$\rho=\frac{0.3534}{3.55\times 10^{-4}}$

$\rho=995.50\ kg.m^{-3}$

Specific weight of the liquid soda:

$\bar w=\frac{m_l.g}{v_l}=\rho.g$

$\bar w=995.5\times 9.81$

$\bar w=9765.887\ N.m^{-3}$

Specific gravity is the density of the substance to the density of water:

$s=\frac{\rho}{\rho_w}$

where:

$\rho_w=$ density of water

$s=\frac{995.5}{1000}$

$s=0.9955$

3 0

3 years ago

Read 2 more answers

If it takes 600 N to move a box 5 meters, how much work is done on the box?

Anika [276]

Answer:

Option C is the correct option.

Explanation:

Given,

Force = 600 N

Distance = 5 meters

Work = ?

Now,

Work = Force $\times$ distance

$= 600 \times 5$

Calculate the product

$= 3000 \:$ Joule

Hope this helps...

Good luck on your assignment..

3 0

3 years ago

Read 2 more answers

When you rub a glass rod with silk. The rod becomes positive charged, and the silk becomes negative.Also when you rub a rubber r

adelina 88 [10]

Answer:

Not sure

Explanation:

7 0

3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$