All categories

2 years ago

Conceptos básicos de la fisica

Physics

1 answer:

Andreyy892 years ago

6 0

Answer:

encontré esto fuera de google, "Estos conceptos físicos que parecen necesarios en cualquier teoría física suficientemente amplia son los llamados conceptos físicos fundamentales, una lista no exhaustiva de los mismos podría ser: espacio, tiempo, energía, masa, carga eléctrica, etc."

espero que esto ayude, que tengas un gran día, Y mantente segura! :) :D :3

You might be interested in

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

2 years ago

What is the weight on the surface of Earth of an object of mass 2.00 kilograms?

Anni [7]

Answer:

Explanation:

19.6 newtons

A 2.00-kilogram object weighs 19.6 newtons on Earth.

6 0

2 years ago

Read 2 more answers

Please help answer question

nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

4 0

2 years ago

If one of two interacting charges is doubled, the force between the charges will _____________.

malfutka [58]

If one of two interacting charges is doubled, the force between the charges will double.

Explanation:

The force between two charges is given by Coulomb's law

$F=\frac{k q1 q2}{r^{2}}$

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Now if the first charge is doubled,

we get $F'=\frac{k (2q1) q2}{r^{2}}$

F'= 2 F

Thus the force gets doubled.

4 0

3 years ago

Acceleration due to gravity is stated as a constant which is given 9.8 m/s2. Many scientists have planned and design experiments

stich3 [128]

The answer is 98.1 m/s2.

3 0

2 years ago