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monitta
2 years ago
6

En que consiste el atletismo de pista y campo?

Physics
1 answer:
OlgaM077 [116]2 years ago
6 0

Answer:

<em>El atletismo es un deporte que incluye competencias atléticas basadas en las habilidades para correr, saltar y lanzar. El nombre se deriva del lugar donde se lleva a cabo el deporte, una pista de atletismo y un campo de hierba para los lanzamientos y algunos de los eventos de salto.</em>

Explanation:

Yo se esto

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Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a
Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg

After that, we compute the potential energy 1.00 m above the reference point:

U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J

Then, we compute the average kinetic energy at the specified temperature:

K=\frac{3}{2}\frac{R}{Na}T

Whereas N_A stands for the Avogadro's number for which we have:

K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0
3 years ago
Aerogenerators are used to produce electricity from
ki77a [65]
Like windmills they use the winds to generate their power.
4 0
3 years ago
wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
Mechanical energy is a term that is used to describe
larisa86 [58]
The sum of potential energy<span> and kinetic </span><span>energy.
Hope I helped!</span>
7 0
3 years ago
The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to
LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power  ,P = F .v

F=force

v=velocity

v= 100 km/h

v=\dfrac{1000}{3600}\times 100\ m/s

v=27.77 m/s

75 x 1000  = F x 27.77

F= \dfrac{75000}{27.77}\ N

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

 a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0
3 years ago
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