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3 years ago

Cells that remain as sources of other cells are called

1 answer:

8 0

They are called stem cells. This cells are undifferentiated which means it can specialize in other types when it receives the right stimuli. They can divide through mitoses and become more stem cell or become a bone, muscle, blood cell, etc.

They can have 2 origins: embryos or some human tissue; their function is to regenerate or substitute damaged cells

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What law of motion is that representing ?

Naya [18.7K]

Explanation:

answer is the 3rd law of motion

I hope it will be helpful for you

please mark as brainest answer

thank you

6 0

3 years ago

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

3 years ago

A piston-cylinder device initially contains 0.3 kg of nitrogen gas at 160 kPa and 140oC. The nitrogen is now expanded isothermal

irakobra [83]

Answer:

<h2>Work done by the gas is given as</h2><h2> $W = 1.72 \times 10^4 J$ </h2>

Explanation:

As we know that the process is isothermal so here work done is given as

$W = nRTln(\frac{P_1}{P_2})$

here we know that

$n = \frac{w}{M}$

$n = \frac{300}{28} = 10.7 moles$

now we have

$W = 10.7 (8.31)(273 + 140) ln(\frac{160}{100})$

so we have

$W = 1.72 \times 10^4 J$

4 0

2 years ago

To get off a frozen lake, a 70 kg person removes his shoe of mass 0.175 kg and throws it horizontally away from the shore at a v

MakcuM [25]

Answer:

the person will be in the shore at 10.73 minutes after launch the shoe.

Explanation:

For this we will use the law of the lineal momentum.

$L_i = L_f$

Also,

L = MV

where M is de mass and V the velocity.

replacing,

$M_i V_i = M_{fp}V_{fp} + M_{fz}V_{fz}$

wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.

so, we will take the direction where be launched the shoe as negative. then:

(70)(0) = (70-0.175)( $V_fp$ ) + (0.175)(-3.2m/s)

solving for $V_fp$ ,

$V_fp$ = $\frac{(3.2)(0.175)}{69.825}$

$V_fp$ = 0.008m/s

for know when the person will be in the shore we will use the rule of three as:

1 second -------------- 0.008m

t seconds-------------- 5.15m

solving for t,

t = 5.15m/0.008m

t = 643.75 seconds = 10.73 minutes

3 0

3 years ago

The law of reflection states that the angle of incidence is equal to the angle of reflection.

djyliett [7]

Yes ... the law of reflection states that the angle of incidence is equal to the angle of reflection <em /><em>
it also states that the incident ray reflected ray and the normal at the point of incidence lie on the same plane....;

</em>

5 0

3 years ago

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