Which force is always attractive and has a very small range?

What causes diffraction that results in a fuzzy glow around a full moon?

bagirrra123 [75]

Well, that's not actually "diffraction".

The fuzzy edge of the moon, and the added glow that's sometimes seen
around it, are all effects caused by the light passing through air before it
reaches you.

This gives you some idea of why astronomers go to such effort and
expense to get their telescopes above as much of the atmosphere as
possible ... placing all serious observatories on mountaintops, and even
putting telescopes in orbit. It's all because the air does such a job on the
light that's trying to shine through it. We have to make do with whatever's
left over after that.

6 0

3 years ago

The net external force on a golf cart is 395.2 n north. if the cart has a total mass of 259.1 kg, what is the cart's acceleratio

yKpoI14uk [10]

Applying the Newton Second Law of motion that is F=ma
We have F=395.2N (North) and m=259.1Kg

Putting these values in the equation

395.2 = 259.1 x a than
for a = 395.2/259.1
a = 1.525 m/sec 2

8 0

3 years ago

Hook's law describes an ideal spring. Many real springs are better described by the restoring force (FSp)s=−kΔs−q(Δs)^3, where q

kvasek [131]

Answer

given,

k = 250 N/m

q = 900 N/m³

(FSp)s=−kΔs−q(Δs)^3

work done = Force x displacement

$W = \int {F. dx}$

limits are x = 0 to x = 0.15 m

work done

$W = \int_0^{0.15} (k x + q x^3)\ dx$

$W = [\dfrac{kx^2}{2}+\dfrac{qx^4}{4}+ C]_0^0.15$

$W = \dfrac{300\times 0.15^2}{2}+\dfrac{900\times 0.15^4}{4}$

W = 3.375 + 0.1139

W = 3.3488 J

b) % cubic term = $\dfrac{0.1139}{3.3488}$

% cubic term = $3.4\ %$

7 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago

Cual es la fuerza electrica sobre el electrón (-1.6 x 10¹⁹c) de un atomo de hidrógeno ejercida por el protón (1.6 x 10¹⁹c)? Supó

kkurt [141]

Answer:

La fuerza eléctrica es -8.2*10⁻⁸ N

Explanation:

El enunciado correcto es: <em>¿Cuál es la fuerza eléctrica sobre el electrón (-1.6 x 10⁻¹⁹c) de un átomo de hidrógeno ejercida por el protón (1.6 x 10⁻¹⁹c)? Supóngase que la distancia entre el electrón y el protón es de 5.3 x 10⁻¹¹ m</em>

Entre dos o más cargas aparece una fuerza denominada fuerza eléctrica. Su valor depende del valor de las cargas y de la distancia que las separa, mientras que su signo depende del signo de cada carga. Las cargas del mismo signo se repelen entre sí, mientras que las de distinto signo se atraen.

La fuerza eléctrica con la que se atraen o repelen dos cargas puntuales en reposo es directamente proporcional al producto de las mismas e inversamente proporcional al cuadrado de la distancia que las separa:

$F=K*\frac{q1*q2}{d^{2} }$

donde:

F es la fuerza eléctrica de atracción o repulsión. En el Sistema Internacional (S.I.) se mide en Newtons (N).
q1 y q2 son lo valores de las dos cargas puntuales. En el S.I. se miden en Culombios (C).
d es el valor de la distancia que las separa. En el S.I. se mide en metros (m).
K es una constante de proporcionalidad llamada constante de la ley de Coulomb. Depende del medio en el que se encuentren las cargas. Para el vacío K tiene un valor aproximadamente de 9*10⁹ $\frac{N*m^{2} }{C^{2} }$ .

En este caso:

F=?

K= 9*10⁹ $\frac{N*m^{2} }{C^{2} }$

q1= -1.6*10⁻¹⁹ C

q2= 1.6*10⁻¹⁹ C

d= 5.3*10⁻¹¹ m

Reemplazando:

$F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{(-1.6*10^{19} C)*(1.6*10^{19} C)}{(5.3*10^{-11} )^{2} }$

Resolviendo:

F= -8.2*10⁻⁸ N

<u><em>La fuerza eléctrica es -8.2*10⁻⁸ N</em></u>

6 0

3 years ago