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notka56 [123]
3 years ago
7

Read the description of the centrioles. What is their function?

Physics
2 answers:
KATRIN_1 [288]3 years ago
7 0

Answer:

Typically found in eukaryotic cells, centrioles are cylindrical (tube-like) structures/organelles composed of microtubules. In the cell, centrioles aid in cell division by facilitating the separation of chromosomes. For this reason, they are located near the nucleus.

Brut [27]3 years ago
3 0

Answer:

sorry im just gettig points but the actaul anserw is on top of this oneExplanation:

Bye

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Sarah and Maisie are analysing data from their school sports day. Looking at the 1500 m results for Stephen, Maisie believes tha
Dahasolnce [82]

Answer:

Sarah is right

Explanation:

This is an exercise that differentiates between scalars and vectors.

A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.

In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero

consequently Sarah is right

4 0
2 years ago
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
3 years ago
Please help I literally don’t understand
Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

7 0
3 years ago
If you answer it I’ll love you forever!!!
Ilia_Sergeevich [38]

Answer:

Performance tests can be used to see if an implemented training program is working for the athlete or if the program needs alterations. They can also assess current abilities in specific athletic areas to help the athlete choose what to focus their energy on improving.

Explanation:

6 0
2 years ago
Read 2 more answers
Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons
yarga [219]

Answer:

q\approx 6.6\cdot 10^{13}~electrons

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

\displaystyle F=k\frac{q^2}{d^2}

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

\displaystyle q=\sqrt{\frac{F}{k}}\cdot d

Substituting values:

\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1

q = 1.05\cdot 10^{-5}~c

This charge corresponds to a number of electrons given by the elementary charge of the electron:

q_e=1.6 \cdot 10^{-19}~c

Thus, the charge of any of the spheres is:

\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}

\mathbf{q\approx 6.6\cdot 10^{13}~electrons}

5 0
2 years ago
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