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torisob [31]
2 years ago
12

People often overestimate the clarity of their intentions in their electronic communications because they underestimate the impo

rtance of ________ in communication.
Physics
1 answer:
maksim [4K]2 years ago
7 0

Answer:

People often overestimate the clarity of their intentions in their electronic communications because they underestimate the importance of tones of voice in communication.

Explanation:

Many factors intervene in communication together. We use words, we use non-verbal language through our gestures and attitudes, and we also use a certain tone of voice. If we focus on this last aspect, the tonality in the voice is decisive in many ways, which cannot be appreciated through electronic communications. With this we communicate emotions, attitudes and a degree of personal involvement. At the same time, the image we project on others largely depends on the tone of voice we have.

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Find the magnitude of the buoyant force on the block. The acceleration of gravity is 9.8 m/s 2 . Answer in units of N.
harkovskaia [24]
The answer is 2248.02 N.
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2 years ago
One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s
DedPeter [7]

Complete Question

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 2 cm across, and you estimate that the distance from the window shade to the wall is about 5 m.

Required:

Estimate the diameter of the pinhole.  

Answer:

The diameter is  d =0.000336 m

Explanation:

     From the question we are told that

            The central maxima is D= 2cm = \frac{2}{100} = 0.02m

            The distance from the window shade is L = 5m

     The  average wavelength of the  sun is mathematically evaluated as

                         \lambda_{ave } = \frac{\lambda_i  + \lambda_f}{2}

 Generally the visible light spectrum  has a wavelength  range  between  400 nm  to 700 nm  

        So  the initial wavelength of the sun is \lambda _i = 400nm

           and the final wavelength is  \lambda_f = 700nm

  Substituting this into the above equation

                 \lambda_{sun} = \frac{400nm  +700nm}{2}

                        = 550nm

The diameter is evaluated as

              d = \frac{2.44 \lambda_{sun} L}{D}

substituting values

              d = \frac{2.44 * 550*10^{-9} * 5 }{0.02}

                d =0.000336 m

5 0
3 years ago
A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o
blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

P=\rho \times g \times h

Here

  • P is the pressure which is given as 68 kPa.
  • ρ is the density of the oil whose SG is 0.92. It is calculated as

                                       \rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\

  • g is the gravitational constant whose value is 9.8 m/s^2
  • h is the height which is to be calculated

                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

Here

  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

                            \Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\

Now corrected pressure is as

P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

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Answer:

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Explanation:

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