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3 years ago

What happens in the gray zone between solid and liquid?-,-

1 answer:

3 0

The gray zone transition is very crucial which includes the inter molecular forces acting on the molecules and each atoms which makes the change in state from hot to cold and cold to hot. and for it to be liquid to solid or solid to liquid the transition needs to cross the gray zone.

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Consider the equation . The dimensions of the variables v, x, and t are , , and , respectively. The numerical factor 3 is dimens

Vinvika [58]

Answer:

Part of the question is missing but here is the equation for the function;

Consider the equation v = (1/3)zxt2. The dimensions of the variables v, x, and t are [L/T], [L], and [T] respectively.

Answer = The dimension for z = 1/T3 i.e 1/ T - raised to power 3

Explanation:

What is applied is the principle of dimensional homogenuity

From the equation V = (1/3)zxt2.

V has a dimension of [L/T]

x has a dimension of [L]

t has a dimension of [T]
from the equation, make z the subject of the relation
z = v/xt2 where 1/3 is treated as a constant
Substituting into the equation for z
z = L/T / L x T2
the dimension for z = 1/T3 i.e 1/ T - raised to power 3

5 0

3 years ago

How many moles of He atoms are in 6.00 g of He?

S_A_V [24]

Answer:

Explanation:

9.5

4 0

3 years ago

2 eggs cook in a frying pan, which is sitting over a gas burner. Is thermal energy being transferred from the metal pan to the e

sesenic [268]

Hope that helps have a good day though.

7 0

3 years ago

Read 2 more answers

Please help u guys acellus sucks

34kurt

Answer:

6.0cm

Explanation:

Given

focal length = 15.0cm

object distance = 10.0cm

Required

Image distance v

Using the formula

1/f = 1/u + 1/v

1/15 = 1/10+1/v

1/v = 1/15 + 1/10

1/v = 2+3/30

1/v = 5/30

v = 30/5

v = 6.0cm

Hence the image distance is 6.0cm

7 0

3 years ago

An electron (charge −e=−1.60×10−19C−e=−1.60×10−19C) is at rest at a distance 1.00×10−10 m 1.00×10−10m from the center of a nucle

shtirl [24]

Answer:

$-9.45\times10^{-16} \text{ J}$

Explanation:

According to Coulomb's law, the force of attraction between two point charges, $q_1$ and $q_2$ , separated by a distance $d$ is given by

$F = k\dfrac{q_1q_2}{d^2}$

$k$ is a constant with a value of $9\times10^9\text{ F/m}$ .

When we substitute the values from the question,

$F = (9\times10^9\text{ F/m})\dfrac{(-1.60\times10^{-19} \text{ C})\times(+1.31\times10^{-17}\text{ C})}{(1.00\times10^{-10}\text{ m})^2} = -1.89\times10^{-6} \text{ N}$

This value is negative because it is in a direction towards the positive charge.

The work done in moving the electron from the nucleus is

$W = F\times r$

$W = (-1.89\times 10^{-6} \text{ N})\times(5.00\times10^{-10}\text{ m}) = -9.45\times10^{-16} \text{ J}$

This is negative because work is done on the electron, not by it.

3 0

3 years ago

What happens in the gray zone between solid and liquid?-,-​

What happens in the gray zone between solid and liquid?-,-