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mr Goodwill [35]
3 years ago
15

A student is given an assignment to demonstrate diffraction. He takes a photograph of a straw in a glass of water. The straw app

ears bent at the water level. Which best describes this example? A)This is a good example of diffraction.B)This is an example of dispersion and not diffraction.C)This is an example of refraction and not diffraction.D)This is an example of reflection and not diffraction.
Physics
2 answers:
morpeh [17]3 years ago
4 0

Diffraction or refraction

Svet_ta [14]3 years ago
4 0

Answer: C) This is an example of refraction and not diffraction.

Explanation: The reason for the straw to appear bent at the water's surface is because of the process of refraction. Light travels at different speeds in different medium because of the varying densities.

The light travels from air to water and thus the difference in speed results in an apparent shift of the position of part of straw under water and thus the part appears broken.

Reflection is the the return of sound or light energy from the surface without absorbing it.

Diffraction is the phenomenon which occurs when the wave encounters an obstacle.

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A 8.01-nC charge is located 1.87 m from a 4.50-nC point charge. (a) Find the magnitude of the electrostatic force that one charg
valkas [14]

Answer:

Explanation:

Given

q_1=8.01\ nC

q_2=4.50\ nC

distance between the charges is given by d=1.87\ m

Electrostatic force between the charges is given by

F_{12}=F_{21}=\frac{kq_1q_2}{d^2}

where k=constant =9\times 10^{9}

F_{12}=F_{21}=\frac{9\times 10^9\times 8.01\times 4.50\times 10^{-18}}{(1.87)^2}

F_{12}=F_{21}=92.769\times 10^{-9}\ N

as both the charges are of similar nature therefore they repel each other

8 0
3 years ago
Extremely confused, please help
Gekata [30.6K]
I think it would be F.
Because the earth is spinning and so is the coin.
7 0
2 years ago
a ball is thrown horizontally from a 20 m high building with a speed of 5.0 m/s. How far from the base of the building does the
kipiarov [429]
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2

Find:
Horizontal displacement

Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s

Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
7 0
3 years ago
Help please! This question is driving me crazy
Varvara68 [4.7K]

Answer:

-10.8°, or 10.8° below the +x axis

Explanation:

The x component of the resultant vector is:

x = 3.14 cos(30.0°) + 2.71 cos(-60.0°)

x = 4.07

The y component of the resultant vector is:

y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)

y = -0.777

Therefore, the angle between the resultant vector and the +x axis is:

θ = atan(y / x)

θ = atan(-0.777 / 4.07)

θ = -10.8°

The angle is -10.8°, or 10.8° below the +x axis.

3 0
3 years ago
A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t
weeeeeb [17]

Answer:

3.59 m/s

Explanation:

We are given that

Mass of lineman,m=85 kg

Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

\theta=90^{\circ}

We have to find the their velocity immediately after the tackle.

Initial momentum,P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s

According to law of conservation of momentum

Initial momentum=Final momentum=(m+m')V

627.6=(85+90)V=175V

V=\frac{627.6}{175}=3.59 m/s

3 0
3 years ago
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