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ser-zykov [4K]
4 years ago
5

A given individual is unable to see objects clearly when he is beyond 60 cm. What power lens should be used to correct this prob

lem? A) -12.932039 D (B) -0.74D (C) -0.828 D (D) -3.529412 D (E) -1.667 Diopters
Physics
1 answer:
Sveta_85 [38]4 years ago
6 0

Answer:

correct option is E

power lens is - 1.667 D

Explanation:

Given data

infinity = 60 cm

to find out

power lens

solution

we know here infinity v = -60 cm because object at infinity

and we take u  = - infinity

so we know that

1/f = 1/v - 1/u

1/f = 1 / -60  - 1 / -infinity

so

1/f = - 1/60

f = - 60 cm = 0.60 m

and we know power of lens = 1/f

so here

power = 1 / -0.60

power = - 1.667 D

so correct option is E

power lens is - 1.667 D

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A 2.40 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N
klemol [59]

Answer: Hello! Here your answer......

Force = 10.244 Newtons

b) No of oscillations = 0.88

Explanation:

Since the block executes SHM we can write it's position as function of time as

ω is the natural frequency of the system

A is the amplitude of the system

Thus accleration of the block

Thus using the given values at t= 3.50 sec we can calculate the acceleration as

thus force can be calculated using newtons second law as

b)

Now no of oscillations can be obtained as

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3 0
3 years ago
On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed
madreJ [45]

Answer:

The length of the rough patch is 4.345 meters.

Explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

K_{1}  = K_{2} + W_{loss}

Where:

K_{1}, K_{2} - Initial and final kinetic energy, measured in joules.

W_{loss} - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

\frac{1}{2}\cdot m \cdot v_{1}^{2}  = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s

Where:

v_{1}, v_{2} - Initial and final speed of the skater, measured in meters per second.

m - Mass of the skater, measured in kilograms.

\Delta s - Length of the rough patch, measured in meters.

f - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

f = r \cdot m \cdot g

Where:

r - Ratio of friction force to weight, dimensionless.

g - Gravitational constant, measured in meters per square second.

Then,

\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s

The equation is simplified algebraically and patch length is cleared afterwards:

\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s

\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }

Given that v_{1} = 5\,\frac{m}{s}, v_{2} = 2.5\,\frac{m}{s}, r = 0.22 and g = 9.807 \,\frac{m}{s^{2}}, the length of the rough patch is:

\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta s = 4.345\,m

The length of the rough patch is 4.345 meters.

6 0
3 years ago
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Answer:

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if it is about electricity then its flagpole

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Explanation:

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Explanation:

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Radioactive atoms are unstable. It decays into stable daughter atom.

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Therefore, Scientists have been able to estimate the age of our solar system by radiometric dating.

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