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3 years ago

What two parts of the eye focus light onto the retina

Physics

1 answer:

melamori03 [73]3 years ago

3 0

Answer: the "cornea" ; and the "lens" .
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Zina [86]

I would say b as well. I’m sorry if it’s wrong

5 0

3 years ago

Select the correct answer.

wolverine [178]

The answer is C in this question.

5 0

3 years ago

Read 2 more answers

Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l

Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m 74 kg

vertical displacement if spider is 11 m

final displacement = 11 cos 60.6 = - 6.753 m

change in displacement is = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is $W = F \delta r cos\theta$

$W = 725.2 \times 4.25 \times cos 180$

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0

3 years ago

The length of the adult mosquito is typically between 3.0 mm and 6.0 mm. The smallest known mosquitoes are around 2.5 mm, and th

FrozenT [24]

0.74in

Explanation:

Given parameters:

length range of adult mosquito = 3.0 - 6.0mm

Smallest mosquito = 2.5mm

Largest mosquito = 19mm

Average mass range = 3.0 - 5.0mg

Unknown:

Length of largest known mosquito in inches = ?

Solution;

The length is the longest dimension. It is how long a body is.

The problem here is converting from mm to inches;

The length of the longest mosquito which is the largest is 19mm

19mm to inches;

1mm = 0.039inches

19mm = 19mm x $\frac{0.039in}{1mm}$ = 0.74in

learn more:

Scale brainly.com/question/570757

#learnwithBrainly

4 0

3 years ago

An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horiz

Dmitry [639]

Answer:

6.3445×10⁻¹⁶ m

Explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

$E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s$

Deflection by Earth's Gravity

$\Delta =\frac {gt^2}{2}$

Now, Time = Distance/Velocity

$\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m$

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m

3 0

3 years ago