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3 years ago

what is the main reason that attitudes are more often revealed in spoken rather than written language

1 answer:

5 0

It's a combination of all those things. probably because we are taught from an early age to write in an academic fashion, giving balanced arguments and a conclusion. When speaking from the heart, there is no opposing argument nor is there a conclusion, just emotion.

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PLEASE HELP!!!

PtichkaEL [24]

Answer:

Densities increase down the group

MP and BP decrease down the group

Softness increased going down the group

Speed of reacting increases going down the group

4 0

3 years ago

The volume of a gas is 605 liters at 27.0°C. The new temperature is -3.0°C. What is the new volume?

nlexa [21]

From p1v1/t1 = p2v2/t2
pressure unchanged ... cancelled out
v1=605 , t1=27C = 300K,
t2=-3C = 270K
***remember temperature must be in Kelvin
we got
605/300 = v2/270
v2 = 545

5 0

3 years ago

Read 2 more answers

Joseph studied whether different materials can block certain electromagnetic waves by testing television reception in different

Sedbober [7]

Joseph's experiment could be improved by using the same antenna at each part of the house during each trial instead of using different antenna. By doing so, he can obtain accurate results how is the signal in different part of the house under the same conditions (despite the location). So, he will see the dependence of the signal on the location. If he uses different antenna, than this antenna can also have influence of the signal.

5 0

4 years ago

The isomerization of cyclopropane to form propene is a first-order reaction. At 760 K, 85% of a sample of cyclopropane changes t

Nataliya [291]

Answer:

so rate constant is 4.00 x 10^-4 $s^{-1}$

Explanation:

Given data

first-order reactions

85% of a sample

changes to propene t = 79.0 min

to find out

rate constant

solution

we know that

first order reaction are

ln [A]/[A]0 = -kt

here [A]0 = 1 and (85%) = 0.85 has change to propene

so that [A] = 1 - 0.85 = 0.15.

that why

[A] / [A]0= 0.15 / 1

[A] / [A]0 = 0.15

here t = (79) × (60s/min) = 4740 s

k = - {ln[A]/[A]0} / t

k = -ln 0.15 / 4740

k = 4.00 x 10^-4 $s^{-1}$

so rate constant is 4.00 x 10^-4 $s^{-1}$

3 0

3 years ago

A bungee jumper of mass m = 65 kg jumps from a bridge.

yawa3891 [41]

To solve the problem it is necessary to apply energy conservation.

By definition we know that kinetic energy is equal to potential energy, therefore

PE = KE

$mgh = \frac{1}{2}mv^2$

Where,

m = mass

g = gravitaty constat

v = velocity

h = height

Re-arrange to find h,

$h=\frac{v^2}{2g}$

Replacing with our values

$h=\frac{26^2}{2*9.8}$

$h = 34.498\approx 34.5m$

Therefore the correct answer is C.

3 0

3 years ago