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zlopas [31]
3 years ago
11

Why does light refract when it encounters the glass in the lens

Physics
2 answers:
marishachu [46]3 years ago
8 0

the answer is B. because it slows down, which causes it to bend

Kobotan [32]3 years ago
5 0
Hey there!

<span>because it slows down, which causes it to bend is the reason why light refracts when it encounters the glass in the lens

Hope this helps!
</span>
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C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F. Calculate the equivalent capacitance
grin007 [14]

The equivalent capacitance between A and B points is 2.5F.

<h3>What is parallel plate capacitor?</h3>

The two parallel plates placed at a distance apart used to store charge when electric supply is on.

The capacitance of a capacitor is given by

C = ε₀ A/d

From the given circuit C1, C2 and C3, C4 are in parallel  C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F

C1, C2 = 4 +4 =8F

C3, C4 = 2 +4 =6F

Now , all capacitors are in series.

Total equivalent capacitance is
1 / Ceq = 1/ 8 +1/6 +1/ 9.2

Ceq = 2.5 F

Thus,  the equivalent capacitance between A and B points is 2.5F.

Learn more about parallel plate capacitor.

brainly.com/question/12733413

#SPJ1

6 0
2 years ago
A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

6 0
3 years ago
A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 18.0t and y =
In-s [12.5K]
Position:                 x = 18t    y = 4t - 4.9t²

First derivative:        x' = 18      y' = 4 - 9.8t

Second derivative:    x'' = 0        y'' = - 9.8


Position vector:      P  =  (18t) i  +  (4t - 4.9t²) j

Velocity vector:      V  =  (18) i  +  (4 - 9.8t) j

Acceleration vector  A  =              (- 9.8) j

6 0
3 years ago
Read 2 more answers
If the electric potential at a point in space is zero, then the electric field at that point is:
icang [17]
E - impossible to determine based on the given information
4 0
3 years ago
Please do all of i will give you brainlest and thanks to best answer
vlabodo [156]

Answer:the answer is A

Explanation:

3 0
3 years ago
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