All categories

3 years ago

What wave property is shown

Physics

2 answers:

pantera1 [17]3 years ago

8 0

Answer:

The answer is Refraction

Lena [83]3 years ago

5 0

Answer:

Refraction.

Explanation:

Refraction is where lighwaves are bent. your eye assumes that the object is where the light ray is coming from when it actually bent to come at your eye from that angle this is why straws appear slightly out of place in water.

You might be interested in

Is the following measurement a vector quantity or a scalar quantity?<br> 68 km South

horsena [70]

This depends if this is Distance or Displacement. If this is distance; Scalar, vector; Vector.

Hope this helps! :)

3 0

3 years ago

Read 2 more answers

What is the electric potential 15 cm above the center of a uniform charge density disk of total charge 10 nC and radius 20 cm?

alisha [4.7K]

Answer:

b) 450 V

Explanation:

We are given that

Total charge, q=10nC= $10\times 10^{-9} C$

$1nC=10^{-9}C$

Radius, r=20 cm= $\frac{20}{100}=0.2m$

1 m=100 cm

x=15 cm=0.15 cm

We have to find the electrical potential 15 cm above the center of a uniform charge density disk .

We know that

$\sigma=\frac{q}{A}=\frac{q}{\pi r^2}$

$\sigma=\frac{10\times10^{-9}}{3.14\times (0.2)^2}$

Where $\pi=3.14$

$\sigma=7.96\times 10^{-8}C/m^2$

Electric potential, $V=\frac{\sigma}{2\epsilon_0}(\sqrt{x^2+r^2}-x)$

Where $\epsilon_0=8.85\times 10^{-12}$

Using the formula

$V=\frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}(\sqrt{(0.15)^2+(0.2)^2}-0.15)$

$V=449.7 V\approx 450V$

Hence, option b is correct.

7 0

2 years ago

Read 2 more answers

In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p

Bond [772]

Answer:

$B_0 = 1.69 \times 10^{-4}\ T$

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

$v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}$

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

$v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}$

v_0 = 6496355.63 m/s

$v_0 = \dfrac{E}{B_0}$

$B_0 = \dfrac{E}{v_0}$

$B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}$

$B_0 = 1.69 \times 10^{-4}\ T$

5 0

3 years ago

What phase difference between two otherwise identical harmonic waves, moving in the same direction along a stretched string, wil

lora16 [44]

Answer:

$\theta=145$

Explanation:

The amplitude of he combined wave is:

$B=2Acos(\theta/2)\\$

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

$0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145$

5 0

3 years ago

The components of vector A are Ax = +2.2 and Ay = -6.9 , and the components of vector B are given are Bx = -6.1 and By = -2.2. W

Zina [86]

For simplicity, let's call vector B-A vector C Then C is
Cx = (-6.1 - 2.2)
Cy = (-2.2 - (-6.9)) Or,
Cx = -8.3 Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = 9.538

3 0

3 years ago