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weeeeeb [17]
3 years ago
12

Una masa de 0,5 kg está sobre una pendiente inclinada 20º sujeta mediante una cuerda paralela a la pendiente que impide que desl

ice. Si no hay rozamiento, ¿qué fuerza hace la cuerda?
Physics
1 answer:
jeka943 years ago
4 0

Answer:

<em>4.61 N</em>

<em></em>

Explanation:

masa = 0.5 kg

ángulo de inclinación = 20°

Peso normal de la masa = mg

donde m = masa

g = aceleración debido a la gravedad = 9.81 m/s^2

Peso normal = 0.5 x 9.81 = 4.905 N

Si la masa se mantiene en su lugar mediante una cuerda paralela al plano, y no hay fricción en la masa, entonces

La fuerza sobre la cuerda = peso normal x cos ∅

donde ∅ = 20 °

La fuerza sobre la cuerda = 4.905 x cos 20°

==><em> 4.61 N</em>

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The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg 


The frequency of mass-spring oscillation is: 
f = (1/2π)√(k/m) 
k = m(2πf)² 

Then we know that k is constant for both trials, we have: 
k = k 


m1(2πf1)² = m2(2πf2)² 

m1 = m2(f2/f1)² 


m1 = (m1+Δm)(f2/f1)² 


m1 = Δm/((f1/f2)²-1)

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5 0
3 years ago
A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni
dimaraw [331]

Answer:

The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})

Put the value into the formula

E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})

E=5.75\ N/C

The direction is toward positive x- axis.

Hence, The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

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2 years ago
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The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
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Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

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Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

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