Wow ! This one could have some twists and turns in it.
Fasten your seat belt. It's going to be a boompy ride.
-- The buoyant force is precisely the missing <em>30N</em> .
-- In order to calculate the density of the frewium sample, we need to know
its mass and its volume. Then, density = mass/volume .
-- From the weight of the sample in air, we can closely calculate its mass.
Weight = (mass) x (gravity)
185N = (mass) x (9.81 m/s²)
Mass = (185N) / (9.81 m/s²) = <u>18.858 kilograms of frewium</u>
-- For its volume, we need to calculate the volume of the displaced water.
The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³. So the volume of the
displaced water (in cm³) is the same as the number of grams in it.
The weight of the displaced water is 30N, and weight = (mass) (gravity).
30N = (mass of the displaced water) x (9.81 m/s²)
Mass = (30N) / (9.81 m/s²) = 3.058 kilograms
Volume of displaced water = <u>3,058 cm³</u>
Finally, density of the frewium sample = (mass)/(volume)
Density = (18,858 grams) / (3,058 cm³) = <em>6.167 gm/cm³</em> (rounded)
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I'm thinking that this must be the hard way to do it,
because I noticed that
(weight in air) / (buoyant force) = 185N / 30N = <u>6.1666...</u>
So apparently . . .
(density of a sample) / (density of water) =
(weight of the sample in air) / (buoyant force in water) .
I never knew that, but it's a good factoid to keep in my tool-box.
Answer:
Distance travelled is 7 meters and the displacement is 3 meters
the answer is d i took the test be cause i go to a k12 onlie school
The P value for the given data set is 25127. For finding P value, we have to must find the Z value.
<h3>How to get the z scores?</h3>
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
The Z value is calculated as;
Z = (X - μ) / σ
Z = (4.007 - 3.6) / 0.607
Z = 0.67051
The P value for the given data set is 25127.
Learn more about z-score here:
brainly.com/question/21262765
#SPJ1
Answer:
It is easier to hear a musician in the classroom than outdoors
Explanation:
It is easier to hear a musician in the classroom due to the improved acoustics provided by the walls of the classroom whereby along with the direct sound of the musician, which is the lead source of the sounds, there is an increased number of indirect sound reaching the ear in the classroom than outdoors and due to precedence effect, all the sound appear to come from the musician
In music played outside, along side the direct sound from the musician, the indirect sound that reach the ear is echoed from maybe by only the ground while the majority of the sound from the music wanders away with the wind and in other directions as well as being absorbed such that speakers will be required to improve the sound of the music outdoors.