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2 years ago

What is the difference between reflection and refraction? What changes and what does not change.

Physics

1 answer:

wolverine [178]2 years ago

8 0

Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

Explanation:

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Question 1<br> 2 pts<br> Explain what causes a solution to be a strong acid.

lubasha [3.4K]

Answer:

Cuanto más fuerte es el ácido, más rápido se disocia para generar H +start superscript, plus, end superscript. Por ejemplo, el ácido clorhídrico (HCl) se disocia completamente en iones hidrógeno y cloruro cuando se mezcla con agua, por lo que se considera un ácido fuerte.

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3 years ago

During an earthquake, you should do all of the following EXCEPT

krok68 [10]

Run inside if you are outdoors .

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4 years ago

What is the kinetic energy of a 478 kg object that is moving with a speed of 15 m/s

Tpy6a [65]

The kinetic energy would be 53,775J:)

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3 years ago

A force (5ỉ - ;)N moves an object from the point P (1,3) m to the point Q (3, 8) m.

natita [175]

Answer:

b) 5 J

Explanation:

Work is the energy transferred by an object when acted by a force along a displacement. Work is the product of force and displacement. The SI unit of work is the joules (J)

To calculate the work done by the force, we have to first get the displacement (D) of the object. Hence:

Displacement (D) = Q(3, 8) - P(1, 3) = (3 - 1, 8 - 3) = (2, 5) = 2i + 5j

The work done is the dot product of the force and the displacement. Force = 5i - j. Hence:

Work done = (5i - j)(2i + 5j) = 10 - 5 = 5 J

3 0

3 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

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3 years ago