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Rama09 [41]
3 years ago
8

Which phenomenon occurs when light falls on a smooth mirror? . . A). Specular reflection. B) diffusion. C) refraction . D) diffu

sed refraction
Physics
2 answers:
gizmo_the_mogwai [7]3 years ago
8 0
Hey there!!!
your answer is going to be a, specular reflection.
hope this helps!!!
SCORPION-xisa [38]3 years ago
3 0
Reflection over a smooth surfaces such as mirrors or water leads to a type of reflection called specular reflection. Reflection over  rough surfaces such as clothing, paper etc. leads to a type of reflection called diffuse reflection<span>.
</span>Therefore,

Specular Reflection occurs when light falls on a smooth mirror.
So the right answer is :
<span>A). Specular reflection.</span>
You might be interested in
Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN ) is a=g(sinθ−μkco
11Alexandr11 [23.1K]

Answer:

a=g(sinθ-μkcosθ)

Explanation:

In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.

wx=m*g*sinθ

wy=m*g*cosθ

As the objects moves down an incline, acceleration in y axis is 0.

Then, by second Newton's Law:

Fy = m*ay

FN - m*g cos θ = 0,

FN=m*g cos θ

In x axis the forces that interacs are the x component of weight and friction force:

Fx = m*ax

mg sen u-FN*μk=m*a

Being friction force, Fr=FN*μk, we replace with its value in below formula:

m*g *sinθ-(m*g*cosθ*μk)=m*a

Then, isolating a:

a=(m*g sinθ-(m*g*cosθ*μk))/m

Solving, we have next equation:

a=g sinθ-(g*cosθ*μk)

Applying distributive property we have:

a=g*(sinθ-μk*cosθ)

5 0
3 years ago
5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley
ella [17]

Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

t = √(1.2/(1/2 ×0.00735)) ≈ 18.07

The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s

5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;

v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

7 0
3 years ago
Read 2 more answers
When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot
gayaneshka [121]

Answer:

The value is E =  1.35 *10^{14} \ J

Explanation:

From the question we are told that

    The mass of matter converted to energy on first test is  m  =  1 \  g  = 0.001 \  kg

    The mass of matter converted to energy on second test m_1 =  1.5 \  g = 1.5 *10^{-3} \ kg

    Generally the amount of energy that was released by  the explosion is  mathematically  represented as  

         E =  m * c^2

=>       E =  1.5 *10^{-3}  * [ 3.0 *10^{8}]^2

=>       E =  1.35 *10^{14} \ J

7 0
3 years ago
A low-luminosity star has a small and narrow ________, whereas a high-luminosity star has a large and wide one.
vova2212 [387]

A low-luminosity star has a small and narrow ​<u>habitable zone</u>, whereas a high-luminosity star has a large and wide one.

<h3>What is luminosity of a star?</h3>

The radiant power emitted by a light-emitting item over time is measured as luminosity, which is an absolute measure of radiated electromagnetic power (light).

The total quantity of electromagnetic energy released per unit of time by a star, galaxy, or other celestial object is referred to as luminosity in astronomy.

Learn more about low-luminosity star:
brainly.com/question/13912549
#SPJ4

4 0
1 year ago
E
N76 [4]

Answer:

Weight

Explanation:

Weight is the downward pull on an object due to gravity.

For example, the moon has less gravity than Earth so we would weigh less on the moon. Our Mass and volume always stay the same but our weight could change.

4 0
4 years ago
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