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N76 [4]
3 years ago
10

Which equation describes the cosine function for a right triangle?

Physics
2 answers:
V125BC [204]3 years ago
6 0

Answer:

cosine is adjacent over hypotenuse

Explanation:

How I know? It's just a basic formula for triangles, and my teacher taught me this. Sorry if this didn't help.

dezoksy [38]3 years ago
3 0

Answer:

D.cos(e)

hypotenuse adjacent

SUENT

Explanation:

it is helpful to you

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explain how potential and kinetic energy are at play when we talk about Newton's second law of motion
KIM [24]

Potential and kinetic energy are at play when we talk about Newton's second law of motion through the various positions in relation to the bodies involved.

<h3>What is Newton's second law of motion?</h3>

This law states that force is equal to the rate of change of momentum and is denoted as F = mv where m is mass and v is velocity.

Potential energy is the energy is possessed by a body by virtue of its position while kinetic energy is possessed by a body by virtue of its motion. Both forms of energy are influenced by forces and are equal to the total momentum.

Read more about Newton's second law of motion here brainly.com/question/2009830

#SPJ1

7 0
1 year ago
NEED HELP ASAP Which of the following chemical equations shows the Law of Conservation of Mass? (1 point)
natulia [17]

Answer:

OK + MgBR arrow KBR + MG

Explanation:

I know nothing about this topic, but if it has to be balanced I am pretty certain that's the only balanced equation

4 0
1 year ago
Hey, what is antimatter
jenyasd209 [6]
A antimatter is particle physics, antimatter is a material composed of the antiparticle "partners" to the corresponding particles of ordinary matter. A particle and its antiparticle have the same mass as one another, but opposite electric charge and other quantum numbers.
5 0
3 years ago
Read 2 more answers
A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w
balu736 [363]

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)

================================================

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.


3 0
3 years ago
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
slega [8]
We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
dI =r^2*dm
where dm =M(b*dr)/(ab)
Now we find the moment of inertia by integrating from -a/2 to a/2
The moment of inertia is
I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3) (from (-a/2) toI=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12 (a/2))



4 0
3 years ago
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