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Yanka [14]
3 years ago
13

Hakeem was dismayed to discover that he offended coworkers by standing 2 feet away from them during business conversations. In t

he U.S., what distance would have been more appropriate? PS 2100
Physics
1 answer:
Eddi Din [679]3 years ago
3 0

Personal space differs from culture to culture, though it is widely acknowledged that Europe and U.S have bigger personal space requirements that their counterparts in Asia.

Hakeem might not realize it but it’s commonly accepted for Americans to have a distance between four to twelve feet between one another in social settings, especially in professional ones. A distance of two feet is only acceptable if the individual is part of the person’s inner circle, such as friends and family.  

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The A string of a violin is a little too tightly stretched. Beats at 4.00 per second are heard when the string is sounded togeth
Basile [38]

Answer:

T=2.5*10^{-3}s

Explanation:

From the question we are told that:

Beat frequency F_b=4

Frequency F=400Hz

Generally the equation for Frequency of the violin is mathematically given by

 f_v=F_b+F

 f_v=4+400Hz

 f_v=404Hz

Therefore the period of the violin string oscillations is

 T=\frac{1}{f_v}

 T=\frac{1}{404}

 T=2.5*10^{-3}s

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3 years ago
The hardest known natural material is _____. <br> jade<br> quartz<br> diamond<br> talc
kakasveta [241]
Its diamond i am 100% right!

8 0
3 years ago
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Calculate the velocity of the bicyclist between 0 and 3 seconds.
hoa [83]

Answer:

is there an equasion it gives you?

Explanation:

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4 0
3 years ago
A running back with a mass of 70 kg travels down the field with a velocity of 5.0 m s . Calculate the kinetic energy of the foot
jarptica [38.1K]

K.E = 1/2 mv²

    =  1/2 (70kg) (5.0ms)²

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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
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