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3 years ago

When a scientist contrasts two or more objects what is he or she looking for?

2 answers:

5 0

When your Contrasting multiple objects, your looking for any differences that may or may not be their. Think Compare and Contrast if that helps, The answer to this is A

brilliants [131]3 years ago

5 0

Answer: a. differences

Explanation:

Making the contrasts means making comparison of the two or more objects on the basis of features they exhibit. A contrast and comparative account will reveal the similarities and differences between the two or more objects. The contrasts can be made on two or more identifiable physical features (such as (size, color, length, height, volume, shape and others) and chemical composition.

A scientist contrasts two or more objects so as to look the differences among them.

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Ohm's law relates the current, voltage, and resistance in a circuit. Use Ohm's law to determine what will happen to the remainin

krek1111 [17]

Answer:

R = ½ R₀

Explanation:

This is an exercise in Ohm's law,

V = IR

in the initial case

V₀ = I₀ R₀ (1)

indicates that the voltage remains constant and the current is doubled

I = 2 I₀

V₀ = I R

we substitute

V₀ = 2 I₀ R

R = ½ V₀ / I₀

we replace by equation 1

R = ½ R₀

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4 years ago

Which science will

PtichkaEL [24]

4) Chemistry .........

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3 years ago

The products of a chemical reaction can never have greater mass than the reactants

jek_recluse [69]

The answer to this is true

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4 years ago

Read 2 more answers

Which one of the following is a decomposition reaction?

dexar [7]

That is not middle school grade level.

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3 years ago

Read 2 more answers

The top of a tower much like the leaning bell tower at Pisa, Italy, moves toward the south at an average rate of 1.4 mm/y. The t

Gemiola [76]

Answer:

$\omega=7.16*10^{-13}\frac{rad}{s}$

Explanation:

The angular speed is given by:

$\omega=\frac{v}{r}$

Here v is the linear speed and r is the radius of the circular motion. The height of the tower is equal to the radius of the circular motion of the top of the tower, since is rotating about its base. We need to convert the given linear speed to $\frac{m}{s}$ :

$1.4\frac{mm}{y}*\frac{10^{-3}m}{1mm}*\frac{1y}{3.154*10^7s}=4.44*10^{-11}\frac{m}{s}$

Now, we calculate the angular speed:

$\omega=\frac{4.44*10^{-11}\frac{m}{s}}{62m}\\\omega=7.16*10^{-13}\frac{rad}{s}$

8 0

4 years ago