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Keith_Richards [23]
3 years ago
10

What is the function of the muscles around the lens in the human eye

Physics
1 answer:
Ede4ka [16]3 years ago
6 0
Its to capture light or to focus. don't forget to like. :D

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A ball rolls from point A to point B. The total energy of the ball at point A isn’t the same as the sum of its potential energy
murzikaleks [220]
B. Some of the ball’s energy is transformed to thermal energy.

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3 0
3 years ago
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Describe three ways you can conserve energy in your own home
Mandarinka [93]
Turn lights off, unplug electronics, and use solar energy
4 0
3 years ago
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What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h
zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s

Induced peak emf = NABω

                               = 500 x (π x 0.25²) x 0.425 x 376.738

                               = 15721.16 V

                               = 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0
3 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
3 years ago
En una fundición hay un horno eléctrico con capacidad para fundir totalmente 540 kg de cobre. Si la temperatura inicial del cobr
11Alexandr11 [23.1K]

Answer:

Q = 2.95*10^5 kJ

Explanation:

In order to calculate the energy required to melt the cooper, you first calculate the energy required to reach the boiling temperature. You use the following formula:

Q_1=mc(T_b-T_1)     (1)

m: mass of cooper = 540 kg

c: specific heat of cooper = 390 J/kg°C

Tb: boiling temperature of cooper = 1080°C

T1: initial temperature of cooper = 20°C

You replace the values of the parameters in the equation (1):

Q_1=(540kg)(390\frac{J}{kg.\°C})(1080\°C-20\°C)=2.23*10^8J

Next, you calculate the energy required to melt the cooper by using the following formula:

Q_2=mL_f         (2)

Lf: melting constant of cooper = 134000J/kg

Q_2=(540kg)(134000\frac{J}{kg})=7.24*10^7J

Finally, the total amount of energy required to melt the cooper from a temperature of 20°C is the sum of Q1 and Q2:

Q=Q_1+Q_2=2.23*10^8J+7.24*10^7J=2.95*10^8J=2.95*10^5kJ

The total energy required is 2.95*10^5 kJ

3 0
3 years ago
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