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2 years ago

15

What is cytoplasm in a animal cell?

1 answer:

qaws [65]2 years ago

6 0

Answer:

Cytoplasm is the jelly-like fluid that fills in all of the. space between the nucleus and the plasma membrane. All of the organelles within a cell are suspended in this fluid. The main function of the cytoplasm is to support the internal structures of the cell as well as maintaining shape and consistency of the cell.

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Julie is cycling at a speed of 3.4 meters/second. If the combined mass of the bicycle and Julie is 30 kilograms, what is the kin

nexus9112 [7]

Answer:

A

Explanation:

KE = 1/2 mv^2

=1/2(30kg)( 3.4 m/s)^2

=173.4 joules

=1.7×10^2 joules

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2 years ago

5. Emanuel Zacchini, the famous human cannonball of the Ringling Bros. and Barnum and Bailey Circus,

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Explanation:

सिद्ध कीजिए किसी भी बराबर भुजाओं वाले त्रिभुज में उनके सामने के कोण बराबर होते है

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2 years ago

An oscillating block-spring system has a mechanical energy of 4.72 J, an amplitude of 11.2 cm, and a maximum speed of 4.24 m/s.

dsp73

Answer:

anyone know this or should i get my brother

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2 years ago

Can velocity and acceleration point in opposite direction?

ICE Princess25 [194]

Yes, therefore the object would then slow down

6 0

3 years ago

Read 2 more answers

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