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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

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 closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc

Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

$f = \dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{1}{1.55}$

$f=0.645\ m$

$f=64.5\ cm$

We need to calculate the image distance

Using lens formula

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}$

$\dfrac{1}{v}=\dfrac{187}{3741}$

$v=20.0\ cm$

Hence, The image distance is 20.0 cm.

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3 years ago

How can you use ecozone in a sentence?

beks73 [17]

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4 years ago

Calculate the force that the 4kg block exerts on the 10kg block

Kryger [21]

Acceleration=force/mass=28/(10+4)=2m/s^2

force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block

vf=vi+at
-10=20*28/14 * t
t=30/2=15sec

i hope this can help you.

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3 years ago

Why is earth the only planet on which water exists in a liquid state

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This is due to earths location in the solar system. Earth is in the habitat zone or the Goldie locks zone, in this zone it's not too hot or not too cold for water to exist. Other planets in different star systems have liquid oceans due to them being in the habitat zone.

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3 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

$V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}$

a).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{t}= 0.7 m^{2}$ , $D_{t}= 0.28$

$F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N$

b).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{h}= 2.44 m^{2}$ , $D_{h}= 0.57$

$F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N$

6 0

3 years ago