I don’t know what to write haha
I don’t know what to write haha - 1

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Answer 1
Answer: I haven’t done this in a while me either but if I knew I would tell u everything I remembered
Answer 2
Answer:

Answer:

words

Explanation:


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Explanation:

Answer:

C

Explanation:

How is carbon monoxide formed in a combustion reaction and what is it?

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Answer:

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What did astronomers name the planet they discovered?

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Calculate the density of a sample of gas with a mass of 30 g and volume of 7500 cm3.

Answers

By definition we have that the density is given by:

Where,

M: mass of the sample

V: volume occupied by the sample

Therefore, substituting values in the given equation we have:

D = 0.004 (g)/(cm^3)

Answer:

the density of a sample of gas with a mass of 30 g and a volume of 7500 cm3 is:

D = 0.004 (g)/(cm^3)

The density of the sample of gas is \boxed{4*10^(-3)\text{ g}/\text{cm}^3} or \boxed{0.004\text{ g}/\text{cm}^3}.  

Further Explanation:

Density of any kind of matter is defined as the total mass of that substance present in the volume occupied by it in the region of space.

Example (1) : If we consider equal amounts of tungsten and iron, then tungsten is heavier than iron because tungsten occupies lesser volume or lesser empty spaces inside it for a particular mass as compared to iron. Therefore, tungsten has density higher than that of iron.

Example (2): If we take 2 cubes of similar size (same volume) of tungsten and gold then the cube of gold will be heavier because its density is more than that of tungsten.

Given:

The mass of the sampleof the gas is 30\text{ g}.

The volume of the sample of the gas is 7500\text{ cm}^3.

Concept:

The density of the substance is defined as the ratio of the mass and volume.

\boxed{\rho=(m)/(V)}                                      ......(1)

Here, \rho is the density of the sample ofthe gas, m is the mass of the sample of the gas and V is the volume occupied by the gas.

Substitute the value of m and V in equation (1).

\begin{aligned}\rho&=\frac{30\text{ g}}{7500\text{ cm}^3}\n&=4*10^(-3)\text{ g}/\text{cm}^3\end{aligned}

Thus, the density of the sample of gas is \boxed{4*10^(-3)\text{ g}/\text{cm}^3} or \boxed{0.004\text{ g}/\text{cm}^3}.

Learn More:

1. Wind and solar energy are the examples of brainly.com/question/1062501

2. Which of the following is not a component of a lever brainly.com/question/1073452

3. A 30kg block being pulled on a carpeted floor brainly.com/question/7031524

Answer Details:

Grade: middle school

Subject: Physics

Chapter: Matter

Keywords:

Density, sample of gas, 30g gas, volume, 7500 cm3, 7500 cm^3, mass, matter, volume occupied, sample, ratio,grams per centimeter cube

PLEASE HELP ME! A uniform rod XY of weight 2.0 N has a length of 80 cm. The rod is suspended by a thread 20 cm from end X. A weight of 5.0 N is suspended from end X. A student hangs a 6.0 N weight on the rod so that it is at equilibrium. What is the distance of the 6.0 N weight from end X.

Answers

Given: Length of rod (L) = 80 cm

Weight of rod (W) =2.0 N

From the attached pictorial diagram,

XY = 80 cm,,XC = 40 cm, XA = 20 cm, AC = 20 cm

Let AB = z cm

Apply the principle of moment about the thread,

            5.0 N × AX = (6.0 N × AB) + (2.0 N ×AC)

or,   5.0 N × 20 cm = (6.0 N × z cm) + (2.0 N ×20 cm)

or.                AB = z = 10 cm

Now, distance XB = XA + AB = 20 cm + 10 cm = 30 cm

Hence, 6.0 N weight should be suspended from X end at distance of 30 cm

Final answer:

6.0 N weight must be placed 0.30 meters (or 30 cm) from end X to keep the rod in equilibrium.

Explanation:

To find the distance of the 6.0 N weight from end X that will keep the rod in equilibrium, you can use the principle of moments (also known as the law of torques). In equilibrium, the sum of the clockwise moments (torques) must be equal to the sum of the counterclockwise moments (torques).

Let's consider the moments (torques) about the point where the rod is suspended:

The 2.0 N weight (the rod's weight) is acting at the midpoint of the rod, which is 40 cm from the suspension point. The moment due to this weight is 2.0 N * 0.40 m = 0.80 N·m in the counterclockwise direction.

The 5.0 N weight at end X is acting at a distance of 20 cm (0.20 m) from the suspension point. The moment due to this weight is 5.0 N * 0.20 m = 1.00 N·m in the counterclockwise direction.

The 6.0 N weight hanging on the rod at an unknown distance "d" from end X creates a moment in the clockwise direction. So, the moment due to this weight is 6.0 N * d m in the clockwise direction.

In equilibrium, the sum of the counterclockwise moments must equal the sum of the clockwise moments:

0.80 N·m + 1.00 N·m = 6.0 N * d m

Now, solve for "d":

1.80 N·m = 6.0 N * d m

Divide both sides by 6.0 N:

d m = 1.80 N·m / 6.0 N = 0.30 m

So, the 6.0 N weight must be placed 0.30 meters (or 30 cm) from end X to keep the rod in equilibrium.

Learn more about Principle of moments here:

brainly.com/question/26117248

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How does hydroelectric energy works?

Answers

Actually, hydroelectric and coal-fired power plants produce electricity in a similar way. In both cases a power source is used to turn a propeller-like piece called a turbine, which then turns a metal shaft in an electric generator,which is the motor that produces electricity. A coal-fired power plant uses steam to turn the turbine blades; whereas a hydroelectric plant uses falling water to turn the turbine. The results are the same. :)