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4 years ago

What can be made in the field or i the laboratory?

1 answer:

ioda4 years ago

6 0

Are there some options?

I think that "observation" is a good option. In the laboratory you can make observations and experiments, but really good controlled experiments are hard in the field... but you can do an observation there!

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How do forces affect the motion of an object?

cricket20 [7]

Explanation:

Unbalanced Forces in Action

Unbalanced forces can change the motion of an object in two ways. ... Second, when unbalanced forces act on a moving object, the velocity of the object will change. Remember that a change in velocity means a change in speed, direction or both speed and direction.

3 0

3 years ago

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

7 0

4 years ago

What is the electrical force between q1 and q2? Recall that k = 8.99 × 109 N•meters squared over Coulombs squared.. 4.3 × 10 N 3

Rudiy27

Answer:

Explanation:

Incomplete question but for understanding.

We want to find the electrical force between two charges, then you can use the coulombs law which states that the force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of their distance apart,

So,

F = kq1•q2 / r²

Where k is a constant and it is given as

K = 8.99 × 10^9 Nm²/C²

q1 and q2 are the charges and in this question it is not given, so the question is incomplete. Let assume that,

q1 = - 1.609 × 10^-19 C electron

q2 = 1.609 × 10^-19 C proton

Since unlike charges attract, then it is force of attraction

Also, r is the distance apart and it is not given, let assume the distance between the two charges is 2 × 10^-5m

Then,

F = kq1•q2 / r²

F = 8.99 × 10^9 × 1.609 × 10^-19 × 1.609 × 10^-19 / (2 × 10^-5)²

F = 5.82 × 10^-19 N

7 0

4 years ago

Read 2 more answers

You are working on a loading dock at UPS. You are instructed by your supervisor to move 20 boxes—each of which has a weight of 1

zhuklara [117]

Answer:

you have patience with me and I will call back tomorrow morning to discuss.

Explanation:

the train leaves and a couple hours of the speech and the other one in my future so I'll have to call them tomorrow to see if you can block them for me happy.

4 0

3 years ago

A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12.0 m/s. What minimum coefficient of friction must

Sergio039 [100]

Explanation:

The centripetal force $F_c$ on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.

$y:\:\:\:\:N - mg = 0$

$x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N$

$m \dfrac{v^2}{r} = \mu mg$

Solving for $\mu$ ,

$\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28$

3 0

3 years ago