All categories

3 years ago

What are five types of organisms That can reproduce asexually

Physics

1 answer:

Sindrei [870]3 years ago

7 0

Answer:

Bacteria, Hydra, Copperheads, Blackworms, and Strawberries

Explanation:

You might be interested in

What is the voltage across a semiconductor bar if the current through it is 0.17 A? The electron concentration in the bar is 2.7

Anastaziya [24]

Answer:

The voltage across a semiconductor bar is 0.068 V.

Explanation:

Given that,

Current = 0.17 A

Electron concentration $n= 2.7\times10^{18}\ cm^{-3}$

Electron mobility $\mu=1000 cm^2/Vs$

Length = 0.1 mm

Area = 500 μm²

We need to calculate the resistivity

Using formula of resistivity

$\sigma=n\times q\times \mu$

$\rho=\dfrac{1}{\sigma}$

Put the value into the formula

$\rho=\dfrac{1}{2.7\times10^{18}\times10^{6}\times1.6\times10^{-19}\times1000\times10^{-4}}$

$\rho=2\ \mu \Omega m$

We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

$R=\dfrac{2\times10^{-6}\times0.1\times10^{-3}}{500\times(10^{-6})^2}$

$R=0.4\ \Omega$

We need to calculate the voltage

Using formula of voltage

$V= IR$

Put the value into the formula

$V=0.17\times0.4$

$V=0.068\ V$

Hence, The voltage across a semiconductor bar is 0.068 V.

6 0

3 years ago

Bro i need help omg.

sergey [27]

Answer:

25.

Explanation:

8 0

2 years ago

Read 2 more answers

A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

$E_{1}+E_{3}-E_{2}=0$

We know that the electric field is:

$E=k\frac{q}{r^{2}}$

Where:

k is the Coulomb constant
q is the charge
r is the distance from the charge to the point

So, we have:

$k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0$

Let's solve it for r(3).

$\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0$

$r_{3}=0.0743\:$

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

3 0

2 years ago

You start pushing a suitcase full of clothes in the horizontal direction with a force of 25.0 newtons. The weight of the suitcas

sasho [114]

Answer:

Distance traveled will be 5.6307 m

Explanation:

Time t = 3 sec

We have given force F = 25 N

We know that force is given by F = ma

So ma = 25 -----------eqn 1

Weight is given by W = 196 N

We know that weight is given by W = mg

So mg = 196 -----------------eqn 2

From equation 1 and equation 2 $\frac{a}{g}=\frac{25}{196}$

$a=1.2512m/sec^2$

Initial velocity is given as 0 so u = 0 m/sec

From second equation of motion $s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 1.2512\times 3^2=5.6307m$

7 0

3 years ago

A ruby laser delivers a 16.0-ns pulse of 4.20-MW average power. If the photons have a wavelength of 694.3 nm, how many are conta

stepan [7]

Answer:

The value is $n = 2.347 *10^{17} \ photons$

Explanation:

From the question we are told that

The amount of power delivered is $P = 4.20 \ M W = 4.20 *10^{6} \ W$

The time taken is $t = 16.0ns = 16.0 *10^{-9} \ s$

The wavelength is $\lambda = 694.3 \ nm = 694.3 *10^{-9} \ m$

Generally the energy delivered is mathematically represented as

$E = P * t = \frac{n * h * c }{\lambda }$

Where $h$ is the Planck's constant with value $h = 6.262 *10^{-34} \ J \cdot s$

c is the speed of light with value $c = 3.0*10^{8} \ m/s$

$4.20 *10^{6} * 16*10^{-9}= \frac{n * 6.626 *10^{-34} * 3.0*10^{8} }{694.3 *10^{-9}}$

=> $n = 2.347 *10^{17} \ photons$

4 0

3 years ago