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Sergeeva-Olga [200]
3 years ago
15

Knowledge and skills learned through socialization are an example of

Physics
1 answer:
ziro4ka [17]3 years ago
3 0

Answer:

I think no.2 the answer

Because socialization and social resources are both for me

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Standard Heat of Formation: The enthalpy change for the formation of 1 mol of a substance in its standard state from its constit
ArbitrLikvidat [17]

Answer:

True

Explanation:

Standard heat of formation is the heat change that deals with the formation of 1mole at standard rates and states of the given reactants . Standard heat of formation is the difference between the enthalpy change of reactants and products.

7 0
3 years ago
Why are we buried in the ground when we die?
trasher [3.6K]
What kind of the question is that. We aren't really buried. We either go to Heaven or the other place. Some people say it's over when your buried in the ground but believers don't really think that. 
3 0
2 years ago
Read 2 more answers
Four blocks of weights are required using which any body whose weight is between 1kg and 40 kg can be weighed. Find the four wei
Vikentia [17]

Answer:

The weights are 1 kg, 3kg, 9kg and 27kg.

Explanation:

The weights are 1 kg, 3kg, 9kg and 27kg.

1+3+9+27= 40

27+9+3= 39

27+9+3-1=38

27+9+1=37

27+9=36

27+9-1=35

27+9+1-3=34

27+9-3=33

27+9-3-1=32

27+3+1=31

27+3=30

27+3-1=29

27+1=28

27

27-1=26

27+1-3=25

27-3=24

27-3-1=23

27+3+1-9=22

27+3-9=21

27+3-9-1=20

Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.

6 0
3 years ago
A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc
marta [7]
This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so: 

v_{d} =  \frac{J}{n\left | q \right |}

<span>The current I is any motion of charge from one region to another, so this is given by:

</span>I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)

The magnitude of the current density is:

J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})

Being:

A=2mm^{2} = 2x10^{-6}m^{2}
<span>
Finally, for the drift velocity magnitude vd, we find:

</span>v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
6 0
3 years ago
Please someone help me.
katovenus [111]

Answer:

acceleration of the car is 3 m\s^2

Explanation:

from rest means the initial velocity (vi) is zero

time = 5s

final velocity (vf) = 15m\s

a = vf - vi \ t

a = (15-0) \ 5

a= 3 m\s^2

which means that the car is speeding up 3 meters every second

5 0
3 years ago
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