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Bess [88]
3 years ago
12

Un movil pasa por el punto A en direccion hacia B (350cm más adelante) y, luego, sigue hasta el punto C. Sabiendo que pasa por B

a las 11:42:38 y por C a las 11:43:16, completando un tiempo total de recorrido de 3min 47s, calule la distancia entre B y C y a la hora a la que paso por el punto A.
Physics
1 answer:
xeze [42]3 years ago
6 0

Explanation:

PRIMERO HACES EL RECUENTO DEL TIEMPO Y LO CONVIERTES EN

SEGUNDOS Y ENTONCES

<em>t</em> = 227 s      t_{AB} = 227 S - 38 s = 189 s

t_{BC} = 38 s

LUEGO USANDO LA ECUACIÓN DE GALILEO GALILEI SSUPONIENDO

QUE EL MOVIL VIAJA A VELOCIDAD CONSTANTE

<em>v</em> = 3.50 m/189 s = 0.0185 m/s

PARA LA DISTANCIA NTRE B Y C

x_{BC} = 0.0185 m/S( 38 s) = 0.703 m

LA HORA EN QUE EL MOVIL PASA POR A ES

11:43:15 - 38 s - 189 s = 11:39:29

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A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm
Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

v =\sqrt{\dfrac{T}{\mu}}

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}

   v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

          λ=2 L

          λ=2 x 0.577 = 1.154 m

we now,

       v = f λ

      f = \dfrac{v}{\lambda}

      f = \dfrac{1074.49}{1.154}

             f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

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