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2 years ago

Identify the processes that are necessary for continued growth and division in a normal body cell

Physics

1 answer:

Vikentia [17]2 years ago

8 0

Answer:

In a normal body cell, the processes that are necessary for continued growth and division is referred to as cell cycle. Cell cycle is the procedure by which cells divide and grow, consisting of specific step by step phases. INTERPHASE is the phase in which the dividing cells spends most of the time resting as it grows to prepare for cell division.

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Kepler’s first law states that:

Likurg_2 [28]

Answer:

Kepler's first law means that planets move around the Sun in elliptical orbits. An ellipse is a shape that resembles a flattened circle. ... It is zero for a perfect circle.

6 0

3 years ago

A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F

ivanzaharov [21]

Answer: $32.65\ m$

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is $\mu_s=0.4$

The coefficient of kinetic friction is $\mu_k=0.2$

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction $\mu_kg=0.2\times 9.8\ m/s^2$

Using the equation of motion $v^2-u^2=2as\\$

insert the values

$0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m$

4 0

3 years ago

What is the symbol name?

murzikaleks [220]

Answer:

cell

Explanation:

here are all the symbols-

hope this helps, take care and stay safe

7 0

3 years ago

Alexandra is attempting to drag her 32.6 kg golden retriever across the wooden floor by applying a horizontal force. What force

vichka [17]

Answer:

230.26 N

Explanation:

Since the speed is constant, acceleration is zero hence the net force will be given by the product of mass, coefficient of friction and acceleration due to gravity

F=0.72*32.6*9.81=230.26 N

7 0

3 years ago

An aluminum cylinder weighing 30 N, 6 cm in diameter and 40 cm long, is falling concentrically through a long vertical sleeve of

Georgia [21]

Answer: Velocity terminal = 0.093m/s

Explanation:

1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)

= (0.0604/2 - 0.06/2)m

= 2×10^-4

Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L

= (π×0.06×0.4)m²

= 0.075m²

Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.

Shearing stress = u×V.terminal/h = 0.86×V/0.0002

= 4300Vterminal

Therefore, Fw = shearing stress × A

30N = 4300Vterminal × 0.075

V. terminal = 30/4300 m.s

V. terminal = 0.093m/s

4 0

3 years ago