Answer:
words
Explanation:
Answer:c is the answer every 2 hours A is just plain no B four hours is too long and D is a no you need to rest
Explanation:
Answer:
C
Explanation:
Answer:
Carbon monoxide is produced when combustion reactions are not fully completed,either through lack of oxygen or due to low mixing.
Carbon monoxide is a colorless and odorless gas.
astronomers named the planet Kepler-186f
Astronomers named the planet they discovered Kepler-186f after the Kepler space telescope, which took the first picture of the planet.
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:
Answer:
the density of a sample of gas with a mass of 30 g and a volume of 7500 cm3 is:
The density of the sample of gas is or .
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 .
The volume of the sample of the gas is .
Concept:
The density of the substance is defined as the ratio of the mass and volume.
......(1)
Here, is the density of the sample ofthe gas, is the mass of the sample of the gas and is the volume occupied by the gas.
Substitute the value of and in equation (1).
Thus, the density of the sample of gas is or .
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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
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
6.0 N weight must be placed 0.30 meters (or 30 cm) from end X to keep the rod in equilibrium.
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.
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