Answer:
=140 cm²
Explanation:
1
____ ×20x 14
2
=1
___ x base × height
is the formula have a gr8 day :)
2
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.
I thinks its He uses proof to show the evidence is relevant. But im not totally positive on it hope this helps
Answer:
Explanation:
Initial velocity u = V₀ in upward direction so it will be negative
u = - V₀
Displacement s = H . It is downwards so it will be positive
Acceleration = g ( positive as it is also downwards )
Using the formula
v² = u² + 2 g s
v² = (- V₀ )² + 2 g H
= V₀² + 2 g H .
v = √ ( V₀² + 2 g H )
A wall uses diffuse reflection while a mirror uses specular reflection. For example, when parallel light rays enter a mirror, they remain parallel when exiting the mirror, allowing you to see a reflection of the light rays. On the contrary, when incident light rays enter a wall which is painted, the rays scatter, not allowing you to see anything but a painted wall.
