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3 years ago

How do scientific tests help determine the properties of substance?

2 answers:

5 0

There are various properties which are exhibited by substances like physical properties,chemical properties,magnetic properties,electric properties etc.

By simply observing the substance we can not get any idea about the property of a substance.One has to conduct various tests in order to reach at his or her conclusion. The scientific tests provides a better understanding about the nature of substance and the changes that happen under various conditions and parameter.

elixir [45]3 years ago

3 0

Scientific tests can help determine substances that appear similar but react differently under the same circumstances. By observing the reactions of the substance using an experiment, you can determine what that substance is made from. Hope this helps!!

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Baseball Digest is but one of several publications on the sport of baseball. True or False?

Gnoma [55]

Answertrue

Explanation:

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3 years ago

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3 years ago

Why is it important to apply critical thinking to spam e-mail?

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A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to

nadya68 [22]

Answer:

Explanation:

Given

Volume of bucket $V=10\ gallon$

Time taken to fill the bucket $t=50\ s$

so volume flow rate is $\dot{V}=\frac{10}{50}=0.2\ gal/s$

1 gal is equivalent to $0.133\ ft^3$

$\dot{V}=0.0267\ ft^3/s$

mass flow rate $\dot{m}=\rho \times \dot{V}$

$\dot{m}=62.4\times 0.0267$

$\dot{m}=1.668\ lbs$

(b)Average velocity through nozzle exit

$\dot{V}=Av_{avg}$

$v_{avg}=\dfrac{0.0267}{\frac{\pi}{4}\times (0.0262)^2}$

$v_{avg}=49.51\ ft/s$

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4 years ago

A Carnot air conditioner operates between an indoor temperature of 20°C and an outdoor temperature of 39°C. How much energy does

Gelneren [198K]

Answer:

D. 130 J

Explanation:

The coefficient of performance for a machine that is being used to cool, is given by:

$COP=\frac{Q_C}{W}=\frac{T_C}{T_H-T_C}$

Here $Q_C$ is the heat removed from the cold reservoir, W is the work required, that is, the energy required to remove the heat from the interior of the house, $T_C$ is the cold temperature and $T_H$ is the hot temperature. Recall use absolutes temperatures( $273.15+^\circ C$ ). Replacing and solving for W:

$W=Q_c\frac{T_H-T_C}{T_C}\\W=2000J\frac{312.15K-293.15K}{293.15K}\\W=129.63J$

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4 years ago