Ask question

Ask question

All categories

2 years ago

15

What is the connection between angle of incidence and angle of reflection

1 answer:

Radda [10]2 years ago

8 0

Answer:

The normal line divides the angle between the incident ray and the reflected ray into two equal angles. The angle between the incident ray and the normal is known as the angle of incidence. The angle between the reflected ray and the normal is known as the angle of reflection.

You might be interested in

The potential in a region of space due to a charge distribution is given by the expression V = ax2z + bxy − cz2 where a = −3.00

Elis [28]

Answer:

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$

Explanation:

We know that the relationship between the electric field $\vec{E}(\vec{r})$ and the potential $V(\vec{r})$ is given by

$\vec{E} ( \vec{r}) = - \vec{\nabla} V(\vec{r})$

So, for our potential:

$V(r) = a x^2 z + b x y - c z^2$

the electric field is :

$\vec{E} ( \vec{r}) = - \vec{\nabla} ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x}, \frac{ \partial }{\partial y} , \frac{ \partial }{\partial z}) ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial y} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial z} ( a x^2 z + b x y - c z^2 ))$

$\vec{E} ( \vec{r}) = - ( 2 a x z + b y , b x , a x^2 - 2 c z )$

This is the our electric field. At vector point

$\vec{r} = (0, -8.00 \ m, - 8.00 \ m)$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 2 a * 0 * (-8.00 \ m) + b (-8.00 \ m) , b * 0 , a 0^2 - 2 c (-8.00 \ m) )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 0 - b 8.00 \ m , 0 , 0 + 2 c 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( b 8.00 \ m , 0 , - 2 c 8.00 \ m )$

Knowing

$b= 6.00 \frac{V}{m^2}$

and

$c=9.00 \frac{V}{m^2}$

the electric field is

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 6.00 \frac{V}{m^2} * 8.00 \ m , 0 , - 9.00 \frac{V}{m^2} * 2* 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$

7 0

2 years ago

A football is thrown with an acceleration of 15 m/s^2 and a force of 13 N. What is its mass?

-Dominant- [34]

I believe it is b. Lmk if I’m wrong

7 0

2 years ago

Read 2 more answers

A 1500 kg car travelling at 25 m/s collides with a 2500 kg van which had

sertanlavr [38]

I’m not sure I think if you google it it should pop up or go on quizlet sorry

6 0

3 years ago

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d

yulyashka [42]

Answer:

a) F = 3.2 10⁻¹⁰ N , b) v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

F = q E

The electric field is related to the potential reference

V = E d

E = V / d

Let's replace

F = e V / d

Let's calculate

F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

v² = v₀ + 2 a d

v = √ 2 ad

Acceleration can be found with Newton's second law

e V / d = m a

a = e / m V / d

a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

v = √ (98,448 10¹⁴)

v = 9.9 10⁷ m / s

3 0

2 years ago

a 8000kg tow truck pulls a 3000kg car horizontally into a garage. if the motor of the tow truck exerts a forward thrust on the t

Evgen [1.6K]

Equation 1 :
m1 : m1a =T
Equation 2:m2 : m2a= F - T

Adding 1 and 2a=0.073

Placing in equation 1

we get T = 218.18N

8 0

3 years ago