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AfilCa [17]
3 years ago
5

A 12.0 cm object is 9.0 cm from a convex mirror that has a focal length of -4.5 cm. What is the distance of the image from the m

irror? –3 cm –4 cm 3 cm 4 cm
Physics
1 answer:
Rasek [7]3 years ago
5 0

Answer:

- 3 cm

Explanation:

From the mirror formula;

1/f = 1/v + 1/u ; where f is the focal length, v is the image distance, and u is the object distance.

1/-4.5 = 1/9 + 1/v

1/v = -1/4.5 - 1/9

    = -1/3

Therefore;

v = -3 cm

Hence;

Image distance is - 3cm

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It bends light rays inwards because it is a converging lense
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Fiora starts riding her bike at 18 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the
lord [1]

Answer:

t = 2.5 hours

Explanation:

given,

speed of the bike for t time= 18 mi/h

final speed of the bike after t time = 12 mi/h

total distance, D = 69 miles

total time, T= 4.5 hour

time for which speed of the bike is 18 mi/h = ?

we know distance = speed x time

now,

18 x t + 12 (4.5 - t) = 69

6 t + 54 = 69

6 t = 15

t = 2.5 hours

The bike was at the speed of 18 mi/h for 2.5 hours.

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The acceleration of a vertically thrown ball at the top of its path is?
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The equation e = mc2 relates mass and <br> a. energy <br> b. force <br> c. work <br> d. gravity
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Una fuerza F de 200 lb actúa a lo largo de AB, sobre la rampa mostrada. fuerza de F respecto del eje OC. Calcule el momento de f
ruslelena [56]

Answer:

Moc = -613.25 [lb*in]

Explanation:

Este problema se puede resolver mediante la mecánica vectorial, es decir se realizara un analisis de vectores.

Primero se calculara el momento de la fuerza F_AB con respecto al punto O, debemos recordar que el momento con respecto a un punto se define como el producto cruz de la distancia por la fuerza.

M_{o}=r_{A/O} * F_{AB} (producto cruz)

Necesitamos identificar los puntos:

O (0,0,0) [in]

A (12,0,0) [in]

B (0, 24,8) [in]

C (12,24,0) [in]

r_{A/O}=(12,0,0) - (0,0,0)\\r_{A/O} = 12 i + 0j+0k [in]\\AB = (0,24,8) - (12,0,0)\\AB = -12i+24j+8k [in]\\[LAB]=\frac{-12i+24j+8k}{\sqrt{(12)^{2} +(24)^{2} +(8)^{2} } }\\ LAB=-\frac{3}{7} i+\frac{6}{7}j+\frac{2}{7}k

El ultimo vector calculado corresponde al vector unitario (magnitud = 1) de AB. El vector fuerza corresponderá al producto del vector unitario por la magnitud de la fuerza = 200 [lb].

F_{AB}=-\frac{600}{7} i +\frac{1200}{7}j+\frac{400}{7} k [Lb]

De esta manera realizando el producto cruz tenemos

M_{O}=r_{A/O} * F_{AB}

M_{O}=0i-685.7j+2057.1k [Lb*in]

Para calcular el momento con respecto a la diagonal OC, necesitamos el vector unitario de esta diagonal.

OC = (12,24,0)-(0,0,0)\\OC= 12i+24j+0k[Lb]\\LOC = \frac{12i+24j+0k}{\sqrt{(12)^{2} +(24)^{2} +(0)^{2} } } \\LOC=\frac{12}{\sqrt{720}}i+\frac{24}{\sqrt{720}}j  +0k

El vector con respecto al eje OC, es igual al producto punto del momento en el punto O por el vector unitario LOC

M_{OC}=L_{OC}*M_{O}\\M_{OC}=(\frac{12}{\sqrt{720}}i +\frac{24}{\sqrt{720}} j+0k )* (0i-685.7j+2057.1k)\\M_{OC}= -613.32[Lb*in]

7 0
3 years ago
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