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Valentin [98]
3 years ago
6

The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point

to 25 cm from this person's eyes?
Physics
1 answer:
makvit [3.9K]3 years ago
5 0
To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m

Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.
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Why is friction important for gymnast working on parallel bars
igomit [66]
Hey there,
Friction is important to increase the friction between the bars and the gymnast to prevent the gymnast from falling and hurting themselves.

Hope this helps :))

<em>~Top♥</em>
8 0
3 years ago
Make the following conversion. 34.9 cL = _____ hL. 0.349 0.0349 0.00349 349,000
AnnyKZ [126]
Hey there!

Here is your answer:

<u><em>The proper answer to this question is option C "</em></u><span><u><em>0.00349".</em></u>

Reason:

</span><span><u><em>1 L = 100 cL. Or 1 cL = 0.01 L</em></u>

</span><span><u><em>34.9 cL = 34.9 / 100 L = 0.349 L</em></u>

</span><span><u><em> 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL</em></u>

<em>Therefore the answer is option C!</em>

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>
4 0
3 years ago
A wire 6.60 m long with diameter of 2.05 mm has a resistance of 0.0310 Ω.
Alex73 [517]

Answer:

1.551×10^-8 Ωm

Explanation:

Resistivity of a material is expressed as shown;.

Resistivity = RA/l

R is the resistance of the material

A is the cross sectional area

l is the length of the wire.

Given;

R = 0.0310 Ω

A = πd²/4

A = π(2.05×10^-3)²/4

A = 0.000013204255/4

A = 0.00000330106375

A = 3.30×10^-6m

l = 6.60m

Substituting this values into the formula for calculating resistivity.

rho = 0.0310× 3.30×10^-6/6.60

rho = 1.023×10^-7/6.60

rho = 1.551×10^-8 Ωm

Hence the resistivity of the material is 1.551×10^-8 Ωm

6 0
3 years ago
A radio wave has a frequency of 5.5 × 104 hertz and travels at a speed of 3.0 × 108 meters/second. What is its wavelength
Ne4ueva [31]
Use v=fλ
3x10^8=5.5x 10^4 λ
λ=5.45x10^3m
4 0
3 years ago
Read 2 more answers
A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I
Anuta_ua [19.1K]

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = \frac{4\times \pi^2\times r^3}{GM} where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

8 0
3 years ago
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