A proton of mass 1.67 x 10-27 kg (the charge of a proton is 1.6 x 10-19 C) enters the region between two parallel plates a
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Answer:
t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s
Explanation:
Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso
y = y₀ + t – ½ g t²
en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)
0 = y₀ – ½ g t²
t= √ (2 y₀/g)
calculemos
t= √ ( 2 12 / 9,8)
t= 1,56 s
El alcance del proyectil es la distancia horizontal recorrida
x = v₀ₓ t
x = 80 1,56
x= 124,8 m
La velocidad de impacto cuando toca el suelo
vx = v₀ₓ = 80 ms
= v_{oy} – gt
v_{y} = - 9,8 1,56
v_{y} = - 15,288 m/s
la velocidad es
v = (80 i^ - 15,288 j ) m/s
Traducttion
This is a projectile launching exercise, let's start by finding the time it takes to reach the ground
y = y₀ + v_{oy} t - ½ g t²
in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( = 0)
0 =y₀I - ½ g t²
t = √ (2y₀ / g)
let's calculate
t = √ (2 12 / 9.8)
t = 1.56 s
Projectile range is the horizontal distance traveled
x = v₀ₓ t
x = 80 1.56
x = 124.8 m
Impact speed when it hits the ground
vₓ = v₀ₓ = 80 ms
v_{y} = v_{oy} - gt
v_{y} = - 9.8 1.56
v_{y} = - 15,288 m / s
the speed is
v = (80 i ^ - 15,288 j) m / s
Answer:
(a) 40.6 degree
Explanation:
When refraction takes place from slab to water, the critical angle is 60 degree.
Use Snell's law
refractive index of water with respect to slab
μs = 1.536
Now for slab air interface, the critical angle is C.
1 / 1.536 = Sin C
C = 40.6 degree
To solve the problem it is necessary to take into account the concepts related to beat frequency, i.e., The number of those wobbles per second.
The equation that describes the beat frequency is
For our given case we have that the frequency of the instrument is 440Hz and the Beat frequency is 5Hz therefore,
A) The frequency of the violin would be given by
B) <em>The violinist must loosen the string.</em> As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i. e, on thightening the string, the frequency further increases as high frequency will be produced by short trings.