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Airida [17]
3 years ago
6

The light incident on a smooth reflecting surface undergoes specular reflection. The angle of reflection is 55°. What can you sa

y about the angle of incidence?
a) It's greater than 55°.
b) It's less than 55°. 
c) It's equal to 55°.
d) It could be any angle between 0° to 90°.
Physics
2 answers:
Serggg [28]3 years ago
7 0
The correct answer is letter C. The angle of incidence is equal to 55 degrees. This is because Specular Reflection refers to mirror-like reflection of light or other kinds of waves from a surface. This phenomenon is where light from a single incoming direction is also reflection in a single-outgoing direction.
Ymorist [56]3 years ago
3 0

Answer: The correct answer is option (c).

Explanation:

According Laws of reflection:

1) The angle of incidence is equal to the angle of reflection, ∠i=∠r.

2)The incident ray , reflected ray and the normal drawn to the surface of the mirror (here smooth reflecting surface) all lies in the sample plane.

So, according to the question a light is reflected back at the angle of 55° which means that angle of the incident ray with the reflection surface is 55° .

Hence the correct answer is option(c).

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Answer:a=2.42 m/s^2

Explanation:

Given

mass m_1=3.50\times 10^{3} kg

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\mu =0.21

Let T be the tension in the rope

From Diagram

m_1g\sin \theta -T-f_r=m_1a-----------------1

where f_r=friction\ force

f_r=\mu m_g\cos \theta

For block m_2

T=m_2a-----------2

From 1 & 2

m_1g\sin \theta -m_2a-\mu m_1g\cos \theta =m_1a

m_1g(\sin \theta -\mu \cos \theta )=(m_1+m_2)a

\frac{9.8\times 3.5\times 10^3}{4.5\times 10^3}(0.5-\0.21\cos 30)=4.5a

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The Problems: 1. Xavier starts at a position of 0 m and moves with an average speed of 0.50 m/s for 3.0 seconds. He normally mov
NemiM [27]

Answer:

(1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

Explanation:

(1). Given that,

Initial position = 0 m

Average speed = 0.50 m/s

Time = 3.0 s

We need to calculate the final position

Using formula of average speed

v_{av}=\dfrac{x_{f}+x_{i}}{t}

Where, x_{f} = final position

x_{i} = Initial position

t = total time

Put the value into the formula

0.50=\dfrac{x_{f}+0}{3.0}

x_{f}=0.50\times3.0

x_{f}=1.5\ m

(2). Given that,

Initial position = 0 m

Average speed = 0.75 m/s

Time = 4.0 s

We need to calculate the final position

Using formula of average speed

v_{av}=\dfrac{x_{f}+x_{i}}{t}

Put the value into the formula

0.75=\dfrac{x_{f}+0}{4.0}

x_{f}=0.75\times4.0

x_{f}=3\ m

(3). Given that,

Average speed = 1.25 m/s

Time = 3.0 sec

Initial position = 1.0 m

We need to calculate the final position

Using formula of average speed

v=\dfrac{x_{f}+x_{i}}{t}

Put the value into the formula

1.25=\dfrac{x_{f}+1.0}{3.0}

x_{f}=1.25\times3.0-1.0

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(4). Given that,

Average speed = 1.25 m/s

Distance = 100 m

We need to calculate the time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{100}{1.25}

t=80 sec

(5). Given that,

Average speed = 5 miles/hr

Suppose, distance = 25 miles

We need to calculate the time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{25}{5}

t=5\ hr

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(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

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