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sukhopar [10]
3 years ago
14

according to the law of reflection what is the relationship between the angle 9f incidence and the angle of reflection

Physics
2 answers:
ryzh [129]3 years ago
7 0

According to the law of reflection, the angle of incidence and the angle of reflection are equal angles.

Alja [10]3 years ago
4 0

Answer:

Angle of incidence is equal to the angle of reflection.

Explanation:

Reflection is defined as the repropagation of light rays incident on a surface. An incident light is a light striking the surface of a plane mirror at an angle to the mirror. This incident light will reflect out at the same angle it strikes the plane. This angle is known as the angle of glance (g). The ray perpendicular to the plane surface is called the normal ray.

Therefore, the angle between the incident ray and the normal ray will be the angle of incidence (i) and the angle between the reflected ray and the normal ray will be the angle of reflection (r). Since the angle the incident ray made with the plane surface is equal to the angle that the reflected ray made with the plane surface then our angle of incidence will also be equal to the angle of reflection according to the second law of reflection i.e i=r.

i = r = 90°-g

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A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
2 years ago
An unstrained horizontal spring has a length of 0.26 m and a spring constant of 180 N/m. Two small charged objects are attached
MakcuM [25]

Answer:

a)

two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)

b)

both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

Explanation:

Given that;

L = 0.26 m

k = 180 N/m

x = 0.039 m

a)

we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.

b)

Spring force F = kx

F = 180 × 0.039

F = 7.02 N

Now, Electrostatic force F = Keq²/r²

where r = L + x = ( 0.26 + 0.039 )

we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2

so from the equation; F = Keq²/r²

Fr² = Keq²

q = √ ( Fr² / Ke )

we substitute

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q =  √ (0.627595 / 9×10⁹)

q = √(6.97 × 10⁻¹¹)

q = 8.35 × 10⁻⁶ C

Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

5 0
2 years ago
Why are triangles important when making structures
Alex_Xolod [135]
So when making structures you have to have squares rectangles right and a triangle and another triangle equals a square so thats all there is to it
4 0
2 years ago
Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
You are watching an object that is moving in shm. when the object is displaced 0.600 m to the right of its equilibrium position,
larisa [96]
It should take at least 4.50

8 0
3 years ago
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