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kupik [55]
3 years ago
12

Label each formula in the chemical quation below as either a reactant of a product.

Physics
1 answer:
Karolina [17]3 years ago
3 0
Fe and S are both reactants: they react with each other to give a different compound.

FeS is the product of the reaction: it was formed or produced as result of the reaction of Fe and S.

Answer:

Fe: reactant
S: reactant
FeS: product


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there is only one basic format for a professional report. a. audience b. All of these should be considered when deciding on a re
Anna007 [38]

Answer:

The answer is letter b. All of these should be considered when deciding on a report format.

Explanation:

A Professional Report is a type of formal document about a topic or information that is intended for a specific audience or purpose. The report's style of writing needs a lot of knowledge from the writer. Oftentimes, it involves the following important elements: <em>Title, Summary, Body, Discussion, Conclusion and Recommendation. </em>

The writer should write according to his target audience and purpose. He also needs to consider the length of his report, as well as the suitable words and sentences that he should use.

Thus, all of the choices are important in writing a professional report. So, the answer is letter b.

5 0
2 years ago
Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J
olchik [2.2K]

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= \sqrt{(8)^2+(-2)^2}

|v|= \sqrt{68} = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

5 0
2 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
At what Fahrenheit temperature are the Celcius and Fahrenheit temperatures numerically the same?
Alex777 [14]

Answer:

-40 degrees

Explanation:

5 0
2 years ago
ANSWER ASAP
Hoochie [10]
The answer to this would inFact be A
6 0
2 years ago
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