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3 years ago

The angle of reflection is the angle the _____ to the reflecting surface.

Physics

2 answers:

worty [1.4K]3 years ago

8 0

Answer:

3. reflected ray makes with a perpendicular line

Explanation:

- The angle of incidence is defined as the angle between the incident ray and the perpendicular to the surface

- The angle of reflection is defined as the angle between the reflected ray and the perpendicular to the surface

According to this second definition, we see that the correct answer that completes the sentence is (3):

The angle of reflection is the angle the 3. reflected ray makes with a perpendicular line to the reflecting surface.

The law of reflection also states that the angle of reflection is equal to the angle of incidence.

Licemer1 [7]3 years ago

5 0

Answer:

reflected ray makes with a perpendicular line

Explanation:

its on my test

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2 years ago

The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. For the same object in low-earth orbit arou

mojhsa [17]

Answer:

g = 8.61 m/s²

Explanation:

distance of the International Space Station form earth is 200 Km

mass of the object = 1 Kg

acceleration due to gravity on earth = 9.8 m/s²

mass of earth = 5.972 x 10²⁴ Kg

acceleration due to gravity = ?

r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n

using formula

$g = \dfrac{GM}{r^2}$

$g = \dfrac{6.67\times 10^{-11}\times 5.972\times 10^24}{(6.8\times 10^6)^2}$

g = 8.61 m/s²

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3 years ago

Name the substance in which starch becomes dark blue in colour.

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Any substance that contains starch turns blue-black in presence of iodine solution.

6 0

2 years ago

A 3.0-kg block moves up a 40o incline with constant speed under the action of a 26-N force acting up and parallel to the incline

CaHeK987 [17]

Answer:

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3 years ago

Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show

Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

$m_2 = \frac{20}{3} kg$

Part b)

Angular acceleration is given as

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = 176.6 N$

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

$m_1g r_1 = m_2 g r_2$

$20 g(0.5) = m_2 g(1.5)$

$10 = 1.5 m_2$

$m_2 = \frac{20}{3} kg$

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

$m_1g - T = m_1 a$

also we have

$T r_1 = I\alpha$

now we have

$m_1g = \frac{I a}{r_1^2} + m_1 a$

$a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}$

$a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}$

$a = 0.981 m/s^2$

so angular acceleration is given as

$\alpha = \frac{a}{r_1}$

$\alpha = \frac{0.981}{0.5}$

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = \frac{I\alpha}{r_1}$

$T = \frac{45 (1.96)}{0.5}$

$T = 176.6 N$

7 0

3 years ago