Can someone help with me 1,2,3 please I will mark brainless .

Un cuerpo de 480 g de masa es atraído con una fuerza de 3.9 E-6 N por otro cuerpo de 196 g de masa. Calcula la distancia a la qu

Soloha48 [4]

Answer:

La distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.

Explanation:

Asumamos que ambos cuerpos son partículas, la fuerza de atracción ( $F$ ), en newtons, entre ambos cuerpos se define mediante la Ley de Newton de la Gravitación Universal, cuya ecuación es:

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

Donde:

$G$ - Constante de la gravitación universal, en metros cúbicos por kilogramo-segundo cuadrado.

$m_{1}, m_{2}$ - Masas de los cuerpos, en kilogramos.

$r$ - Distancia entre los cuerpos, en metros.

Si sabemos que $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $m_{1} = 0.48\,kg$ , $m_{2} = 0.196\,kg$ y $F = 3.9\times 10^{-6}\,N$ , entonces la distancia entre los dos cuerpos es:

$r = \sqrt{\frac{G\cdot m_{1}\cdot m_{2}}{F} }$

$r \approx 1.269\times 10^{-3}\,m$

Es decir, la distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.

5 0

3 years ago

Chapter 38, Problem 001

Norma-Jean [14]

Answer:

a) $\lambda=2.95x10^{-6}m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=\frac{c}{\lambda}$

If we replace the last equation into the E formula we got:

$E=h\frac{c}{\lambda}$

And if we solve for $\lambda$ we got:

$\lambda =\frac{hc}{E}$

Using the value of the constant $h=4.136x10^{-15} eVs$ we have this:

$\lambda=\frac{4.136x10^{15}eVs (3x10^8 \frac{m}{s})}{0.42eV}=2.95x10^{-6}m$

$\lambda=2.95x10^{-6}m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^{-6}m$

4 0

3 years ago

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of static friction between t

Wewaii [24]

Answer:

a) a = 6.1 m/s^2

b) a = 0.98m/s^2

Explanation:

Mass of slab = 40kg

Mass of block = 10kg

Coefficient of static friction (Us) = 0.60

Kinetic coefficient (UK) = 0.40

Horizontal force = 100N

The normal reaction from 40kg slab on 10 kg block = 10*9.81

= 98.1N

Static frictional force = Us*R

= 98.1*0.6

= 58.86N

This is less than the force applied

If 10 kg block will slide on the 40 kg slab, net force = 100 - kinetic force

Kinetic force (Uk*R) = 0.4*98.1

= 39.28N

= 39N

Net force = 100 -39

= 61N

Recall that F = ma

For 10 kg block

a = F/m

a = 61/10

a = 6.1m/s^2

b) Frictional force on 40 kg slab by 10 kg = 98.1*0.4

= 39.24

= 39N

F = ma

a = F/m

For 40kg slab

a = 39/40

a = 0.98m/s^2

3 0

3 years ago

Why can a bowling ball do work

Karo-lina-s [1.5K]

it cant beacuse its not a person

8 0

3 years ago

A closed pipe creates a

lina2011 [118]

Answer:

375 hz

Explanation:

Trust me

5 0

3 years ago