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2 years ago

Fred wants to summarize mitosis in the cell cycle. Which statement describes mitosis?

2 answers:

6 0

The statement that most accurately describes mitosis simply is <span>that mitosis is a type of cell division in which one cell (the mother) divides to produce two new cells (the daughters) that are genetically identical to itself</span>. This is the most basic textbook definition, that is summary, of what the mitosis is.

antiseptic1488 [7]2 years ago

6 0

the answer is Mitosis includes several steps that occur after the DNA has been copied but before the cytoplasm is divided

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Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as

Sonja [21]

Explanation:

The waveform expression is given by :

$y(x,t)=y_m\ sin(0.333x+5.36+585t)$ ...........(1)

Where

y is the position

t is the time in seconds

The general waveform equation is given by :

$y(x,t)=y_m\ sin(kx+\phi+\omega t)$ ..........(2)

Where

$k=\dfrac{2\pi}{\lambda}$

$\omega=2\pi f$

On comparing equation (1) and (2) we get :

$0.333=\dfrac{2\pi}{\lambda}$

$\lambda=18.86\ m$

$585=2\pi f$

f = 93.10 Hz

Time period, $T=\dfrac{1}{f}$

$T=\dfrac{1}{0.010}$

T = 0.010 s

Phase constant, $\phi=5.36\ radian$

Hence, this is the required solution.

8 0

2 years ago

an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and

natulia [17]

Answer:

The skater's speed after she stops pushing on the wall is 1.745 m/s.

Explanation:

Given that,

The average force exerted on the wall by an ice skater, F = 120 N

Time, t = 0.8 seconds

Mass of the skater, m = 55 kg

It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

$\Delta p=m(v-u)\\\\\Delta p=mv$

The change in momentum is equal to the impulse delivered. So,

$J=\Delta p=F\times t\\\\mv=F\times t\\\\v=\dfrac{Ft}{m}\\\\v=\dfrac{120\times 0.8}{55}\\\\v=1.745\ m/s$

So, the skater's speed after she stops pushing on the wall is 1.745 m/s.

4 0

1 year ago

A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t

Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, $a_t$ = 12.7 m/s²

mass of the truck, $m_t$ = 2490 kg

mass of the car, $m_c$ = 890 kg

let the acceleration of the car at the moment they collided = $a_c$

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

$F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2$

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0

2 years ago

A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a

Yanka [14]

Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

Explanation:

We have given young's modulus of steel $E=207GPa=207\times 10^9Pa$

Change in length $\Delta l=0.38mm$

Length of rod $l=500mm$

Load F = 11100 KN

Strain is given by $strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}$

We know that young's modulus $E=\frac{stress}{strain}$

So $207\times 10^9=\frac{stress}{7.6\times 10^{-4}}$

$stress=1573.2\times 10^{-5}N/m^2$

We know that stress $=\frac{force}{artea }$

So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

$area=7.055\times 10^{8}m^2$

So $\frac{\pi }{4}d^2=7.055\times 10^{8}$

$d=2.998\times 10^4m$

6 0

2 years ago

Given that the CFL bulb is rated at 10% efficiency, if the LED bulb consumes half the amount of electrical power, for the same a

Butoxors [25]

Answer and explanation:

The efficiency of the LED bulb is 20% while The efficiency of the Inc bulb is 1%.

The lightbulb efficiency can be defined as the quotient between the amount of usable light and the amount of electrical power output. For the efficiency of the CFL Bulb:

$\epsilon_{CFL}=\frac{L_0}{P_0}=0.1$

Therefore:

$\epsilon_{led}=\frac{L_0}{0.5P_0}=2 \cdot0.1=0.2\\\epsilon_{INC}=\frac{L_0}{10P_0}=0.1 \cdot 0.1=0.01$

4 0

1 year ago