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3 years ago

6

HURRY DUE TODAY!!! Most animals begin their lives as a single, microscopic cell called a zygote. This begins a series of countle

ss cell divisions, eventually resulting in a large, complex organism. Which type of cell division is used for this process?
A.
conjugation
B.
meiosis
C.
mitosis
D.
budding

2 answers:

BigorU [14]3 years ago

6 0

Answer:

I think B or C but I don't know

posledela3 years ago

6 0

The answer:

C) mitosis

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A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answe

Elina [12.6K]

A) +5 J

B) +1 J

Explanation:

A)

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

$\Delta U=-5 J$ is the loss in potential energy of the system

$\Delta K=+6 J$ is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

B)

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

$E=K+U$

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

$\Delta E= \Delta K + \Delta U$

In this problem:

$\Delta K=+6 J$

$\Delta U=-5 J$

So, the change in mechanical energy is:

$\Delta E=+6+(-5)=+1 J$

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

$W=\Delta E=+1 J$

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4 years ago

If a display has a dynamic range of 20 dB and the smallest voltage it can handle is 200 mV, then the largest voltage it can hand

DedPeter [7]

Answer:

The largest voltage is 0.02 V.

(d) is correct option.

Explanation:

Given that,

Range = 20 dB

Smallest voltage = 200 mV

We need to calculate the largest voltage

Using formula of voltage gain

$G_{dB}=10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$20 =10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$2=log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$10^2=\dfrac{V_{out}^2}{V_{in}^{2}}$

$\dfrac{V_{out}}{V_{in}^}=10$

$V_{in}=\dfrac{V_{out}}{10}$

$V_{in}=\dfrac{200}{10}$

$V_{in}=20\ mV$

$V_{in}=0.02\ V$

Hence, The largest voltage is 0.02 V.

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3 years ago

Why a pond of water appears shallower than it actually is?

GalinKa [24]

The refraction of light at the surface of water makes ponds look shallower then they really are

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3 years ago

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago

What is <br> 0.02204 in scientific notation

Studentka2010 [4]

Answer:

2.204x10^2 -2 = 0.02204 in scientific notation.

4 0

3 years ago