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Oliga [24]
2 years ago
7

Hans observed properties of four different waves and recorded observations about each one in his chart.

Physics
2 answers:
Andre45 [30]2 years ago
7 0
D.<span>Wave 3 resulted from constructive interference, and Wave 4 resulted from destructive interference.

</span>
Hatshy [7]2 years ago
7 0

i think it is d, because wave w is being diffracted wave x is being reflected, and wave y and z are being refracted.

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To eight significant figures, Avogadro's constant is 6.0221367×10^(23)mol−1. Which of the following choices demonstrates correct
Zepler [3.9K]

Answer with Explanation:

We are given Avogadro's constant =6.0221367\times 10^{23}mol^{-1}

There are eight significant figures.

We have to round off.

1.If we round off to four significant figures

The ten thousandth place of Avogadro's constant is less than five therefore, digits on left side of ten thousandth  place remains same and digits on right side of ten thousandth place and ten thousandth place  replace by zero.

 Then ,Avogadro's constant can be written as

6.022\times 10^{23}mol^{-1}

If we round off to 2 significant figures

Hundredth place of given number is less than 5 therefore, digits on left side of hundredth  place remains same and digits on right side of hundredth place and hundredth place replace by zero.

Then,Avogadro's constant can be written as

6.0\times 10^{23}mol^{-1}

If we round off six significant figures

6 is greater than 5 therefore, 1 will be added to 3 and digits on right side of 6 and 6 replace by zero and digits on left side of 6 remains same except 3.

Then, the Avogadro's constant can be written as

6.02214\times 10^{23}mol^{-1}

3 0
2 years ago
Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi
AleksandrR [38]
The horizontal component is calculated as:
Vhorizontal = V · cos(angle)

In your case Vhoriontal = 16 · cos(40) = 12.3 m/s

Answer: 12.3 m/s
7 0
2 years ago
Read 2 more answers
A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the
frozen [14]

Answer:

8.0\cdot 10^5 N/C

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

E=k\frac{q}{(R+r)^2}

where

k=8.99\cdot 10^9 N m^2C^{-2} is the Coulomb's constant

q=2.0 \mu C=2.0 \cdot 10^{-6}C is the charge on the sphere

R=10 cm = 0.10 m is the radius of the sphere

r=5.0 cm = 0.05 m is the distance from the surface of the sphere

Substituting, we find

E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C

3 0
2 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
2 years ago
A company designed and sells an ultrasonic​ receiver, which detects sounds unable to be heard by the human ear. The receiver can
Ksenya-84 [330]

Answer:

28√3 m

Explanation:

A = vertex where receiver is placed

S = focus

Bp = r = radius of the outside edge

Bc = 2r = diameter

The full explanation is shown in the picture attached herewith. Thank you and i hope it helps.

7 0
2 years ago
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