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Katen [24]
3 years ago
10

An object with a mass of 20 kg has a net force of 80 N acting on it. What is the acceleration of the object?

Physics
2 answers:
Anastaziya [24]3 years ago
5 0

Answer:

acceleration  =  4 m/s²

Explanation:

alexdok [17]3 years ago
4 0
Newton's 2nd law of motion says

                   Net force = (mass) x (acceleration)

Plug in the things you know, and you have

                   80 N  =  (20 kg) x (acceleration)

Divide each side by (20 kg) :

                  80N / 20kg  =  acceleration

                   acceleration  =  4 m/s²
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katrin2010 [14]

Answer:

Energy is applied on the charge to do work.

Explanation:

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3 years ago
The graph that BEST shows the relationship between colour and wavelength is​
Lady bird [3.3K]
Answer is A I hope it helps
6 0
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[C]utting programs that were not effective or essential and even some that were, but are now unaffordable; and precisely targeti
zalisa [80]

Option B i.e discretion is the right and correct answer.

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The President above clearly refers to the Discretion Spending where portion of the budget is decided by Congress through the annual appropriations.

3 0
3 years ago
Read 2 more answers
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
4 years ago
Consider the possibility of using two rotating cylinders to replace the conventional wings on an airplane for lift. Consider an
EleoNora [17]

Answer:

27.35m

Explanation:

For the calculation of the Support Force we rely on the formula for obtaining the force in a cylinder of a certain length l,

F_y = - \rho Ul\Gamma

Here each term is,

F_y= Lift force

\rho= density of air

\Gamma = vortex strength

For this last equation, its mathematical representation is given by,

\Gamma = 2\pi av_{\theta}

Here each term is,

a= 1m, radios of cylinder

v_{\theta}= 20 Km/hr=5.5m/s, the velocity of cylinder surface.

\Gamma = 2\pi (1)(5.5) = 34.90m^2/s

In order to find the density of the area at 2000m we will refer to the table of Standard Atmosphere of the United States, that is 1.007kg/m^3,

U= 150Km/hr = 41.6m/s, F_y = 40000N, \Gamma = 34.90m^2/s

Replacing the values,

40000 = -(1.007)(41.6)l(34.90)

Clearing l and solving for it we have,

l=-27.35m

<em>In this way we can conclude that the length of the cylinder must be 27.35m</em>

7 0
4 years ago
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