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3 years ago

Does light slow down or speed up when it passes from the air into the cornea? How do you know this?

1 answer:

3 0

Light slows down when it passes from the air to the cornea because the cornea supply’s two thirds of the power of the eye. The speed of light changes significantly when traveling from air into the cornea.

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A mule is harnessed to a sled having a mass of 246 kg, including supplies. The mule must exert a force exceeding 1210 N at an an

zimovet [89]

(a) 1800 N

The equation of the forces along the vertical direction is:

$F sin \theta + N - mg = 0$

where

$F sin \theta$ is the component of the applied force along the vertical direction

N is the normal force on the sled

mg is the weight of the sled

Substituting:

F = 1210 N

m = 246 kg

$\theta = 30.3^{\circ}$

We find N:

$N=mg-F sin \theta = (246)(9.8)-(1210)(sin 30.3^{\circ})=1800 N$

(b) 0.580

The equation of the forces along the horizontal direction is:

$F cos \theta - \mu_s N = 0$

where

$F cos \theta$ is the horizontal component of the push applied by the mule

$\mu_s N$ is the static frictional force

Substituting:

F = 1210 N

N = 1800 N

$\theta = 30.3^{\circ}$

We find $\mu_s$ , the coefficient of static friction:

$\mu_s = \frac{F cos \theta}{N}=\frac{(1210)(cos 30.3^{\circ})}{1800}=0.580$

(c) 522 N

In this case, the force exerted by the mule is

$F= 6.05 \cdot 10^2 N = 605 N$

So now the equation of the forces along the horizontal direction can be written as

$F cos \theta - F_f = 0$

where

$\theta=30.3^{\circ}$

and $F_f$ is the new frictional force, which is different from part (b) (because the value of the force of friction ranges from zero to the maximum value $\mu N$ , depending on how much force is applied in the opposite direction)

Solving the equation,

$F_f = F cos \theta = (605)(cos 30.3^{\circ})=522 N$

6 0

3 years ago

Joe decides to go bungee jumping. He secures a bungee to his ankles and leaps off a tall bridge. At one location, for one instan

slavikrds [6]

Gravitational energy i believe

6 0

3 years ago

A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on

kari74 [83]

Answer:

$9.16rad/s^2$

Explanation:

We are given that

Mass, $m_1=3.3 kg$

Radius,r=0.8 m

$m_2=4.9 kg$

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

$4.9g-T=4.9a$

$Tr=I\alpha$

Where

$\alpha=\frac{a}{r}$

$Tr=\frac{1}{2}m_1ra$

$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

$4.9g-\frac{1}{2}(3.3a)=4.9a$

$4.9\times 9.8=4.9a+\frac{3.3a}{2}$

Where $g=9.8 m/s^2$

$48.02=a(4.9+1.65)=6.55a$

$a=\frac{48.02}{6.55}=7.33m/s^2$

Angular acceleration, $\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2$

7 0

3 years ago

One of two nonconducting spherical shells of radius a carries a charge Q uniformly distributed over its surface, the other carri

lisabon 2012 [21]

Answer:

Explanation:

A ) The spheres are non conducting , charge will not move on the surface so neutralization of charge by + ve and - ve charge is not possible. Charges will remain intact on them . The electric field inside them will be zero . Electric field outside shell will not be spherically symmetrical . Lines of force will emanate from the surface of positively charged shell outwardly oriented and end at negatively charged shell .

B )

distance between the centres of spherical shell

= 2 a

potential energy of charges

= k q₁ x q₂ / R

= k x - Q x Q / ( 2a )

= - k Q²/ 2a

So work needed to separate them to infinity will be equal to

= k Q²/ 2a

6 0

3 years ago

The phase of matter which is exposed to normal atmospheric pressure is solely dependent upon temperature? true or false

Diano4ka-milaya [45]

The statement above is true. The phase of matter which is exposed to normal atmospheric pressure is indeed solely dependent upon temperature. If the matter is exposed to the normal atm pressure, its temperature depends on it.

6 0

3 years ago