2 years ago

What is the answer to this question?

Physics

1 answer:

solmaris [256]2 years ago

8 0

Answer:

NA2SO3 is the reactant in the reaction

Explanation:

The substances which take part in a chemical reaction are called reactants

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A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.

DENIUS [597]

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy $Q_{out}=1000\ Btu$

Work done $W_{cycle}=300\ Btu$

We need to calculate the coefficient of performance

Using formula of the coefficient of performance

$COP=\dftrac{Q_{in}}{W_{cycle}}$

We need to calculate the $Q_{in}$

$W_{cycle}=Q_{out}-Q_{in}$

Put the value into the formula

$300=1000-Q_{in}$

$Q_{in}=300-1000$

$Q_{in}=700\ Btu$

Now put the value of $Q_{in}$ into the formula of COP

$COP=\dfrac{700}{300}$

$COP=\dfrac{7}{3}=2.33$

Hence, The coefficient of performance for the cycle is 2.33.

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3 years ago

The pressure of a gas is 15 atm in a 5L cylinder. If the volume of the cylinder is depressed to 3L, What is the new pressure exe

disa [49]

P1v1=p2v2. 15x5=p2x3

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3 years ago

The electric eye that prevents an elevator door from closing when you are in the doorway relies upon which physics principle?

goblinko [34]

It's Photoelectric Effect, I just a test with this same question. I am not good for explaining exactly how, but I was right.

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3 years ago

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A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement?

Lorico [155]

20miles (16^2+12^2)^1/2=20

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3 years ago

An archer pulls her bowstring back 0.396 m by exerting a force that increases uniformly from zero to 237 N. (a) What is the equi

tatyana61 [14]

Answer:

(a) The equivalent spring constant is 598.485 N/m

(b) The work done is 46.926 J

Explanation:

From Hooke's law of elasticity

K (spring constant) = F/e

F is the range of force exerted = 237 - 0 = 237 N

e is the extension of bowstring = 0.396 m

K = F/e = 237/0.396 = 598.485 N/m

Work done = 1/2 Fe = 1/2 × 237 × 0.396 = 46.926 J

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3 years ago