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Nookie1986 [14]
3 years ago
13

What is the function of eye lens of human eye​

Physics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

the lens is located in the eye. by changing its shape, the lens changes the focal distance of the eye. in other words,it focuses the light rays that pass through it (and onto the retina) in order to create clear images of objects that are positioned

at various distances.

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In a swimming meet, the swimmers swim a total of 8 laps of a 50-meter-long swimming pool. What is the distance traveled by a swi
N76 [4]

Answer:

The swimmer has a distance traveled of 800 meters.

The final displacement of the swimmer is 0 meters.

Explanation:

A lap is a round trip made by a swimmer in the pool, so that the distance traveled by swimmer is sixteen times the length of the swimming pool. That is:

s = \left(2\,\frac{travels}{lap} \right)\cdot \left(8\,laps \right)\cdot \left(50\,\frac{m}{travel}\right)

s = 800\,m

A swimmer has a distance traveled of 800 meters.

The displacement is the distance between swimmer and a reference point, let suppose that reference point is located at the beginning of the first lap. Hence, the final displacement of the swimmer is 0 meters.

7 0
2 years ago
A proton is observed to have an instantaneous acceleration 11*10^11. what is the magnitude of e of the electric field at the pro
jek_recluse [69]

The magnitude of the electric field at the proton's location is 10,437.5 N/C.

<h3>What the magnitude of the electric field?</h3>

The size of the electric field is basically characterized as the power per charge on the test charge. On the off chance that the electric field strength is meant by the image E. Very much like gravity, electric fields work the same way. In any case, while gravity generally draws in, an electric field, then again, can either rebuff or draw in. By and large, the Electric Field submits to the super-position guideline. the all out Electric Field from various charges is equivalent to the amount of the electric fields from each charge separately. An electric field is the actual field that encompasses electrically charged particles and applies force on any remaining charged particles in the field, either drawing in or repulsing them.

Learn more about the magnitude of the electric field, visit

brainly.com/question/26898699

#SPJ4

6 0
2 years ago
Two objects gravitationally attract with a force of 10N. If the distance between the two objects center is doubled, then the new
lana [24]

Answer:

20N

Explanation:

10×2

5 0
3 years ago
Most geologists believe that the dinosaurs became extinct 65 million years ago when a large comet or asteroid struck the earth,
Alex17521 [72]

Answer:

A) 1.67 x 10 ⁻⁶ m/s

B)5.59 x 10^-^9 %

Explanation:

A)

Given:

d = 5.0 km,

mₐ = 2.5 x 10^1^4 kg

u₁ = 4.0 x 10⁴ m/s

m_n = 5.98 x 10 ²⁴ kg

Solve using kinetic conserved energy

mₐ x u₁ + m_n  x u₂ = uₓ x (mₐ + m_n )

(2.5 x 10^1^4) (4.0 x 10⁴ )+ (5.98 x 10 ²⁴ )(0) = uₓ x (2.5 x 10^1^4 + 5.98 x 10 ²⁴ )

uₓ = ( 2.5 x 10^1^4 x 4.0 x 10⁴ ) / (2.5 x 10^1^4 + 5.98 x 10 ²⁴ )    

uₓ = 1.67 x 10 ⁻⁶ m/s

B) Assuming earth radius as a R = 1.5 x 10 ¹¹ m

t = 365 days  x 24 hr / 1 day x 60 minute / 1 hr x 60s / 1 minute = 31536000 s

t = 31536000 s

D = 2 π R = 2 π( 1.5 x 10 ¹¹ )

D = 9.4247 x 10 ¹¹ m

u₂  = D / t  = 9.4247 x 10 ¹¹  / 31536000

u₂ =  29885.775 m/s

% = (  1.67 x 10 ⁻⁶ m/s ) / (29885.775 m/s) x 100

% = 5.59 x 10^-^9 %

5 0
3 years ago
The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou
Cerrena [4.2K]

Answer:

a) I = 19.799\,kg\cdot m^{2}, b) T = -3.405\,N\cdot m, c) n_{T} \approx 54.842\,rev

Explanation:

a) The net torque is:

T = I\cdot \alpha

Let assume a constant angular acceleration, which is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}

\alpha = 1.793\,\frac{rad}{s^{2}}

The moment of inertia of the wheel is:

I = \frac{T}{\alpha}

I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }

I = 19.799\,kg\cdot m^{2}

b) The deceleration of the wheel is due to the friction force. The deceleration is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}

\alpha = - 0.172\,\frac{rad}{s^{2}}

The magnitude of the torque due to friction:

T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )

T = -3.405\,N\cdot m

c) The total angular displacement is:

\theta_{T} = \theta_{1} + \theta_{2}

\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}

\theta_{T} = 344.580\,rad

The total number of revolutions of the wheel is:

n_{T} = \frac{\theta_{T}}{2\pi}

n_{T} = \frac{344.580\,rad}{2\pi}

n_{T} \approx 54.842\,rev

5 0
3 years ago
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