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Phantasy [73]
3 years ago
6

A person looks across a stadium at the scoreboard, which appears blurry. The person looks down at the program, which is complete

ly in focus. What vision condition is described?
a) color blindness

b) astigmatism

c) farsightedness

d) nearsightedness
Physics
2 answers:
Alecsey [184]3 years ago
5 0

Answer:

d) nearsightedness

Explanation:

Nearsightedness (myopia) is a common vision condition in which you can see objects near to you clearly, but objects farther away are blurry. It occurs when the shape of your eye causes light rays to bend (refract) incorrectly, focusing images in front of your retina instead of on your retina

Pepsi [2]3 years ago
3 0
Answer is D. Nearsightedness is when a person can see near, but not far. Everything appears blurry from far away, but as you get close to it, it becomes more focused
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What is the net magnetic flux through any closed surface?
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Answer:

The net magnetic flux through any closed surface must always be zero.

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3 years ago
The heating element in a kettle behaves like a resistor. A particular kettle needs to operate at 230 V, with a power of 1500 W.
dedylja [7]

Answer:

R = 35.27 Ohms

Explanation:

Given the following data;

Voltage = 230V

Power = 1500W

To find the resistance, R;

Power = V²/R

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R is the resistance measured in ohms.

Substituting into the equation, we have;

1500 = 230²/R

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R = 35.27 Ohms.

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4 0
2 years ago
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

6 0
3 years ago
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