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tatuchka [14]
2 years ago
14

Two converging lenses are separated by 27.0 cm. The focal length of each lens is 8.90 cm. An object is placed 33.0 cm to the lef

t of the lens that is on the left. Determine the final image distance relative to the lens on the right.
Physics
1 answer:
Ymorist [56]2 years ago
7 0

Answer:

1/i + 1/o = 1/f     thin lens equation

i = 33 * 8.9 / (33 - 8.9) = 12.2 cm  to right of first lens

27 - 12.2 = 14.8 cm to left of second lens

i = 14.8 * 8.9 / (14.8 - 8.9) = 22,3 cm to right of second lens

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Answer:

C

Explanation:

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6 0
3 years ago
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shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at const
Contact [7]

Answer:

a) T=0.40 N

b) T=1.9 s

Explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

P=2\pi r=0.94

r=\frac{0.94}{2\pi}=0.15 m

Now, we need to find the angle of the pendulum from vertical.

tan(\alpha)=\frac{r}{L}=\frac{0.15}{0.90}=0.17

\alpha=9.44 ^{\circ}

Let's apply Newton's second law to find the tension.

\sum F=ma_{c}=m\omega^{2}r

We use centripetal acceleration here, because we have a circular motion.

The vertical equation of motion will be:

Tcos(\alpha)=mg (1)

The horizontal equation of motion will be:

Tsin(\alpha)=m\omega^{2}r (2)

a) We can find T usinf the equation (1):

T=\frac {mg}{cos(\alpha)}=\frac{0.04*9.81}{cos(9.44)}=0.40 N

We can find the angular velocity (ω) from the equation (2):

\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s

b) We know that the period is T=2π/ω, therefore:

T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s

I hope it helps you!

8 0
3 years ago
How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?
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8 0
3 years ago
A lion has a mass of 45 kg. Answer the following questions about it, using correct units. a. The lion runs at a speed of 14.2 m/
Eva8 [605]
A) The kinetic energy of an object is given by:
K= \frac{1}{2}mv^2
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K= \frac{1}{2}mv^2= \frac{1}{2}(45 kg)(14.2 m/s)^2=4537 J

b) the increase in gravitational potential energy of the lion is given by:
\Delta U = mg \Delta h
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Kaylis [27]
The orbital with the lowest energy is 3s.
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