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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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sladkih [1.3K]

Answer:

d=1.29*10^{-6}m

Explanation:

From the question we are told that:

Distance of wall from CD D=1.4

Second bright fringe y_2= 0.803 m

Let

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Generally the equation for Interference is mathematically given by

y=frac{n*\lambda*D}{d}

Where

d=\frac{n*\lambda*D}{y}

d=\frac{2*431 *10^{-9}m*1.4}{0.803}

d=1.29*10^{-6}m

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2 years ago
Graphs are representations of equations.<br> A. True<br> B. False
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Answer:

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3 years ago
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A cat, walking along the window ledge of a new york apartment, knocks off a flower pot, which falls to the street 280 feet below
Harlamova29_29 [7]
H = 280 ft, the height of the flower pot.
g = 32 ft/s²

Neglect air resistance.
Note that 1 ft/s = 15/22 mi/h

The initial vertical velocity is zero.
Let v =  the velocity with which the flower pot hits the ground.
Then
v² = 2gh
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v = 133.866 ft/s

Also,
v = (133.866 ft/s)*(15/22 (mi/h)/(ft/s)) = 91.272 mi/h

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5 0
3 years ago
A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T
slamgirl [31]

Answer:

u=14.48m/s

Explanation:

From the question we are told that:

Height of window h=2m

Height of window off the ground h_g=7.5m

Time to fall and drop t=1.3s

 

Generally the Newton's equation motion  is mathematically given by

 s=ut+\frac{1}{2}at^2

Where

h=ut+\frac{1}{2}at^2

2=u1.3-\frac{1}{2}*9.8*1.3^2

2=u1.3-8.281

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Generally the Newton's equation motion  is mathematically given by

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Where

-2gh_g=v^2-u^2

-2*9.8*7.5=(7.91)^2-u^2

-147=62.5681-u^2

u=\sqrt{209.5681}

u=14.48m/s

Therefore the  ball’s initial speed

u=14.48m/s

8 0
3 years ago
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