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4 years ago

Which part of the eye controls whether or not your pupil dilates or restricts?

Physics

1 answer:

valentinak56 [21]4 years ago

5 0

The answer would be, the iris. Hope this helps:)

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A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?

Lyrx [107]

W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.

3 0

3 years ago

Read 2 more answers

A laser beam of wavelength 600 nm is incident on two slits that are separated by 0.02 mm. What is the separation between adjacen

Liula [17]

Answer:

option D

Explanation:

given,

wavelength = 600 nm

width of separation = 0.02 mm

L = 5 m

for mth order maxima

$d \times \dfrac{y_m}{L}=m\lambda$

for (m+1)th order maxima

$d \times \dfrac{y_{m+1}}{L}=(m+1)\lambda$

now,

$y_m=\dfrac{mL\lambda}{d}$ and

$y_{m+1}=\dfrac{(m+1)L\lambda}{d}$

hence,

$\Delta y = y_{m+1} - y_m$

$\Delta y =\dfrac{L\lambda}{d}$

$\Delta y =\dfrac{5 \times 600 \times 10^{-9}}{0.02 \times 10^{-3}}$

$\Delta y =0.15\ m$

$\Delta y =15\ cm$

hence, the correct answer is option D

4 0

4 years ago

Flag question

anastassius [24]

Answer:

P = density (p) * g * h

P = 1000 kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2

since kg m / s^2 = Newtons

The average pressure is 1/2 (pressure at 0m + pressure 80 m) for liquid of uniform density

3 0

3 years ago

Which material will displace a volume of water

Nat2105 [25]

Its probably a graduated cylinder of something in that nature

3 0

3 years ago

Read 2 more answers

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the

miskamm [114]

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

$V =\frac{Kq}{r}$

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V$

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V$

Total potential due to this charges = 4500 V + 6750 V = 11250 V

3 0

4 years ago