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2 years ago

Why are a 12 ounce hard seltzer and 1.5 ounces of liquor both standard drinks? alchohol edu

1 answer:

6 0

12 ounce hard seltzer and 1.5 ounces of liquor are standard drinks because they contain the <u>same amount</u><u> </u><u>of </u><u>pure alcohol</u>

<h3>What is standard drink?</h3>

This is a term used to refer to the measure of alcoholic content of drinks such that the drink should have 14 grams of pure alcohol or 0.6 fluid ounces of pure alcohol. This concept of standard drink is applicable in the United States of America.

The equivalence of a standard drink is 5 percent alcohol as seen in regular beer, 12 percent as seen in wines, and 40 percent as seen in distilled spirits. This measurement in taken by the percentage of the total volume of the beverage. hence the quantity of the drink or beverage may be different as the percentage is what determines the amount of alcohol present

From standard drink chart both drinks has same amount of pure alcohol so they are said to be standard drinks

Read more on standard drinks here: brainly.com/question/17645986

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Does a neutrally charged object even have a charge?

Roman55 [17]

“Charged objects have an imbalance of charge - either more negative electrons than positive protons or vice versa. And neutral objects have a balance of charge - equal numbers of protons and electrons. The principle stated earlier for atoms can be applied to objects. Objects with more electrons than protons are charged negatively; objects with fewer electrons than protons are charged positively.

In this discussion of electrically charged versus electrically neutral objects, the neutron has been neglected. Neutrons, being electrically neutral play no role in this unit. Their presence (or absence) will have no direct bearing upon whether an object is charged or uncharged. Their role in the atom is merely to provide stability to the nucleus.”

Hope this helps a bit.

!! (Credits to The Psychics Classroom) !!

8 0

3 years ago

Read 2 more answers

The normal force which the path exerts on a particle is always perpendicular to the _________________ tangent to the path. trans

Marianna [84]

Tangent to the pathh.

7 0

3 years ago

A missile is launched vertically from a missile silo, it will explode after 32 s. It's launch speed was 145 m/s, and it doesn't

Karolina [17]

Answer:

Landed before it explodes

Explanation:

vf = vi + at,

0 = 145 - (9.8)t,

t = 14.79 s (Time to reach highest point)

14.79 x 2 = 29.59 s (Time to land on the ground)

It will have landed before it explodes because both the time to reach the highest point and the time to land on the ground are less than 32 seconds.

4 0

2 years ago

A chemical reaction takes place in which energy is absorbed. Arrange the characteristics of the reaction in order from start to

Ira Lisetskai [31]

<h2>Answer</h2>

It will be single step endothermic reaction.

<h2>Expalantion</h2>

In the endothermic reaction, the reactants come together to convert to products by absorbing the heat from the external source. This reaction is explained is also known as one step reaction as reactants meet to get the transition stage and converts to the product. But in some reactions, the activation energy required to activate the reactants to get the transition stage to form products. For this, the reaction will have the steps as activation energy, reactant meet, transition stage and products form.

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3 years ago

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6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced: