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3 years ago

State the law of refraction

Physics

2 answers:

IgorLugansk [536]3 years ago

8 0

Answer:

The law of refraction states that the incident ray, the refracted ray, and the normal to the interface, all lie in the same plane.

Explanation:

Vlad1618 [11]3 years ago

8 0

Answer:

1 ) angle of incidence is always equal to the angle of reflection

2) the incidence ray , the normal at the point of incidence and the reflection ray , all lie in the same plane

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An object is moving with a constant velocity of 278 m/s. How long will it take it to travel 7500 m, using the formula Delta X=Vt

rusak2 [61]

If the velocity is constant then the acceleration of the object is zero.
$a=0 (m/s^2)$
Thus when we apply the equation
$\Delta X=vt+(at^2/2)$
It remains
$\Delta X =vt$
or equivalent
$t=(\Delta X/v) =7500/278 =26.98 (seconds)$

7 0

3 years ago

In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau

mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

Em₀ = U = m g y₁

final point. At the exit of the ramp

Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

Em₀ = Em_f

m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

v = w r

the moment of inertia of a sphere is

I = $\frac{2}{5}$ m r²

we substitute

m g (y₁ - y₂) = ½ m v² + ½ ( $\frac{2}{5}$ m r²) ( $\frac{v}{r}$ )²

m g h = ½ m v² (1 + $\frac{2}{5}$ )

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

mg h = $\frac{7}{5}$ ( $\frac{1}{2}$ m v²)

v = √5/7 √2gh

This is the exit velocity of the vertical movement of the sphere

v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

starting point

Em₀ = U = mg y₁

final point

Em_f = K + U = ½ m v² + m g y₂

Em₀ = Em_f

mg y₁ = ½ m v² + m g y₂

m g (y₁ -y₂) = ½ m v²

v = √2gh

this is the speed of the box

v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

v² = v₀² - 2 g y

at the highest point v = 0

y = vo₀²/ 2g

for the sphere

y_sphere = 5/7 2gh / 2g

y_esfera = 5/7 h

for the block

y_block = 2gh / 2g

y_block = h

therefore the block reaches higher than the sphere

$\frac{y_{sphere}} {y_bolck} = 5/7$

3 0

3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

2 years ago

Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ = 4.0 × 10-6 C

Len [333]

Answer: 71.93 *10^3 N/C

Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:

∫E*dr=Q inside/εo Q inside is given by: λ*L then,

E*2*π*r*L=λ*L/εo

E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C

6 0

3 years ago

A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location

amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

$f=-16.0 cm$ (the focal length is negative for a diverging lens)

$p=10.0 cm$ is the distance of the object from the lens

Solvign the equation for q, we find

$\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}$

$q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm$

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0

3 years ago