What is the medium for this wave?

castortr0y [4]

No. I do not agree with Stefan. Quite the contrary. I disagree
with his description of "<span>angle of incidence" as the angle between
the surface of the mirror and the incoming ray.

The correct description of "angle of incidence" is </span><span>the angle between
the NORMAL TO the surface of the mirror and the incoming ray.

Thus, the true angle of incidence is the complement of the angle that
Stefan calculates or measures.</span>

5 0

3 years ago

Read 2 more answers

A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet

astraxan [27]

Answer:

Acceleration = 2.35 m/ $s^{2}$

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = $\frac{mg}{2}$ = 0.5 mg

Friction along incline = umg cos30° = mg $0.3\times \frac{\sqrt{3}}{2}$

Friction along incline = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = $\frac{net force}{mass}$ = 0.24 g = 2.35 m/ $s^{2}$

The height of incline = 8 m

Length of the inclined edge = 16 m

$v^{2}=u^{2}+2as$

$v^{2}= 2\times 0.24 \times 9.8\times 16$

v= 8.67 m/s

5 0

3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

2 years ago

A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W

dezoksy [38]

Answer:

Explanation:

a ) Slit separation d = .1 x 10⁻³ m

Screen distance D = 4 m

wave length of light λ = 650 x 10⁻⁹ m

Width of central fringe = λ D / d

= $\frac{650\times10^{-9}\times4}{.1\times10^{-3}}$

= 26 mm

b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm

c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to

λ / d

= $\frac{650\times10^{-9}}{.1\times10^{-3}}$

= 6.5 x 10⁻³ radian.

8 0

3 years ago

What is the temperature 32 degrees Fahrenheit in degrees Celsius? A. 0°C B. 20°C C. 10°C D. –10°C

omeli [17]

The answer would 0. The reasoning of this is because freezing point in celsius is always 0 degrees but in fahrenheit the freezing point is 32 degrees.

8 0

3 years ago