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katrin [286]
3 years ago
6

An object is placed 100 cm in front of a diverging lens of focal length -25cm. A converging lens of focal length 33 1/3 cm is pl

aced 30 cm past the first lens. What is the lateral magnification of this system of lenses?
Physics
1 answer:
bagirrra123 [75]3 years ago
8 0
We use 1/o + 1/i = 1/f  where o is the distance of the object, i as distance of the image and f is the focal length.
Substituting, <span>1/ 100 + 1 / i = - 1 /25 </span>
<span>i = - 20 cm </span>

<span>For the case of the problem,</span>

<span>o = (20 + 30)  = 50 cm </span>

<span>f = 33.33. </span>Using 1<span> / i + 1 / o = 1/f , </span><span> </span><span>i = 100 cm </span>

<span>M = magnification = - i / o </span>

<span>m1 = -(-20)/100 = 20/100 = 0.2 </span>

<span>m2 = -100/50 = -2 </span>

<span>M = m1*m2 = -2 x 0.2 = -0.4.</span>
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If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.

The resultant of the two forces F1 and F2 is
F_{12} =  \sqrt{F_1^2+F_2^2}= \sqrt{(10 N)^2+(10 N)^2}= \sqrt{200}=14.1 N
with direction at 45^{\circ} north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be 45^{\circ} south-east.
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3 years ago
A driver of a car enters a new 110 km/h speed zone on the highway. The driver begins to accelerate immediately and reaches 110 k
levacccp [35]

Answer:

30Km/h

Explanation:

acceleration is the change of speed in a given time so when we substract the accelerations we can know how much the car goes per an hour

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Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual
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Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

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2 years ago
Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in
USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>

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b) We will calculate the force for a speed of 30 m/s.

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