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3 years ago

The image shows water passing through a barrier.

Physics

2 answers:

Aloiza [94]3 years ago

8 0

C is the answer for this question

zloy xaker [14]3 years ago

3 0

It is diffraction

Explanation:

The opening is the aperture

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A motorcycle traveling at 25 m/s accelerates at a rate of 6 m/s2 for 4 seconds. What is the final speed of the motorcycle in m/s

givi [52]

Initial velocity = Vo= 25 m/s

Final velocity = V = x

Acceleration= a = 6 m/s^2

time= t = 4 seconds

Appy the equation:

V = Vo + at

Replacing:

V = 25 + 6(4) = 25 + 24 = 49 m/s

8 0

10 months ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

2 years ago

If a person is 6.25 m away from a 60.0 w speaker, what is the sound level they are hearing?

vodka [1.7K]

Answer:

110.87 dB

Explanation:

(I got it right on Acellus)

I= P/4(pi)r^2 = 60/4(pi)6.25^2

60/4(pi)6.25^2=0.12223

B=10log(I/Io)

B=10log(0.12223/1*10^-12) = 110.87 dB

111 in sigfigs

4 0

2 years ago

Read 2 more answers

A real power supply can be modeled as an ideal EMF of 60 Volts in series with an internal resistance. The voltage across the ter

lara31 [8.8K]

Answer:

5 ohms

Explanation:

Given:

EMF of the ideal battery (E) = 60 V

Voltage across the terminals of the battery (V) = 40 V

Current across the terminals (I) = 4 A

Let the internal resistance be 'r'.

Now, we know that, the voltage drop in the battery is given as:

$V_d=Ir$

Therefore, the voltage across the terminals of the battery is given as:

$V= E-V_d\\\\V=E-Ir$

Now, rewriting in terms of 'r', we get:

$Ir=E-V\\\\r=\frac{E-V}{I}$

Plug in the given values and solve for 'r'. This gives,

$r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms$

Therefore, the internal resistance of the battery is 5 ohms.

5 0

3 years ago

Following are the different layers of the sun’s atmosphere. Rank them based on the order in which a probe would encounter them w

blsea [12.9K]

Answer:

Going from earth to the sun a probe would encounter the next layers in order:

Corona
Transition Region
Chromosphere
Photosphere
Convection Zone
Radiative Zone
Core

A brief description of them:

Corona is the outermost layer and it cannot be seen with the naked eye, is starts at about 2100 km from the surface of the sun and it has no limit defined.

Transition Region is between the corona and the chromosphere, it has an extension of about 100km

The chromosphere is between 400 km from the surface of the sun to 2100 km. In this layer the further you get away from the sun it gets hotter.

The photosphere is the surface of the sun, the part that we can see, and extends from the surface to 400km.

The convection zone is where convection happens, hot gas rises, cools and rises again.

Radiative Zone is where the photons try to rise to move to higher layers.

The core of the Sun is where nuclear fusion occurs due to the very high temperatures.

6 0

3 years ago