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SSSSS [86.1K]
2 years ago
7

In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the

south wall, sitting 0.780 m away from it. To enable her to see the choir, a flat mirror 0.600 m wide is mounted on the south wall, straight in front of her. What width of the north wall can the organist see?
Physics
1 answer:
ololo11 [35]2 years ago
7 0

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

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A proton moves from a location where V = 87 V to a spot where V = -40 V. (a) What is the change in the proton's kinetic energy?
Art [367]

Answer: a) 127 eV; b) there is no change of kinetic energy.

Explanation: In order to explain this problem we have to use the change of potentail energy ( conservative field) is equal to changes in kinetic energy. So for the proton ther move to lower potential then they gain kinetic energy from the electric field.  This means the electric force do work in this trayectory and then the protons increased changes its speed.

If we replace the proton by a electron we have a very different situaction, the electrons are located in a lower potental then  they can not move to higher potential  if any  external force does work on the system.

In resumem, the electrons do not move from a point with V=87 to other point with V=-40 V. The electric force point to high potential so the electrons  can not move to lower potential region (V=-40V).

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3 years ago
What is the wavelength of a 6.00*10^2 Hz sound wave in air at 20C
Alex Ar [27]

Answer:

Solution

λ=v/n

Here, v=344 m s−1

n=22 MHz =22×106 Hz

λ=344/22×106=15.64×10−6m=15.64μm.

5 0
2 years ago
A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm
andrew11 [14]

Answer:

2.8 cm

Explanation:

y_1 = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm

Fringe width is also given by

\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}

For second order

\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1

Distance between two second order minima is given by

y_2=2\beta_2

\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm

The distance between the two second order minima is 2.8 cm

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Ca anyone help me with writing a lab report about the reaction between the pepper and the soap in water
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Answer:

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Explanation:

Surface tension:

The surface tension of a liquid is the tendency of liquid surfaces to resist an external force due to the cohesive nature of its molecules.

The pepper and soap experiment helps you to understand buoyancy force and surface tension.

Reaction between the pepper and soap is as following.

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This is the reaction that take places between soap and pepper experiment.

Learn more about Pepper and soap experiment here:

<u>brainly.com/question/9614070</u>

<u>#SPJ4</u>

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