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2 years ago

What are 2 laws of reflection

Physics

2 answers:

densk [106]2 years ago

8 0

Answer:

The two laws of reflection are as follows: ... The incidenct ray, the reflected ray, and the normal to the reflecting interface at the point of incidence all lie in the same plane. 2. The angle of incidence equals the angle of reflection.

Shalnov [3]2 years ago

5 0

Angle of incidence and Angle of reflection hope this helps

You might be interested in

What component of a longitudinal sound wave is analogous to a trough of a transverse wave?

anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.

6 0

2 years ago

Short Answer Questions:

Otrada [13]

According to Archimede's principle, a physical object experiences an upthrust due to a difference in pressure between upper and lower surfaces.

<h3>What is an upthrust?</h3>

An upthrust is also referred to as buoyancy and it can be defined as an upward force which is exerted by a fluid (liquid or gas), so as to oppose the weight of a partially or fully immersed physical object that is floating in it.

Based on scientific information, a physical object experiences an upthrust when it is immersed in a fluid due to a difference in height and pressure between upper (top) and lower (bottom) surfaces.

According to Archimede's principle, there is a higher pressure at the bottom of the physical object due to height, and a lower pressure at the top of the physical object.

Read more on upthrust here: brainly.com/question/24389514

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4 0

1 year ago

A total of 875 J of work was done when a force of 175 N was exerted on a box to move it. How far was the box moved

Vladimir [108]

Work = Force times Distance
W = Fd

Given W = 750J, F = 125N;

750 = 125d

Solving for d:
d = 750/125
d = 6

The box moved a distance of 6 meters.

3 0

2 years ago

A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed

Tresset [83]

For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg

4 0

2 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.