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3 years ago

What is another term for a Lewis structure diagram?

Physics

1 answer:

Natasha2012 [34]3 years ago

8 0

Answer:

electron dot diagram

Explanation:

i took the test

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Four hundred seventy joules of heat are removed from a heat reservoir at a temperature of 470 K. What is the entropy change of t

11111nata11111 [884]

Answer:

ΔS= -1 J/K

Explanation:

Given data

Heat Q= -470J

Temperature T=470 K

To find

Entropy change ΔS

Solution

We know that the entropy change of system is ΔS is given by

ΔS=Q/T

We have take heat value Q as negative because the heat is removed from heat reservoir

ΔS=(-470J/470K)

ΔS= -1 J/K

8 0

3 years ago

Read 2 more answers

Gravity on earth is 9.8m/s^2, and gravity on the moon is 1.6 m/s^2. So if the mass of an object on earth is 40 kilograms, it’s m

tigry1 [53]

31.6 kilograms mass on the

7 0

4 years ago

Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and tot

Oksi-84 [34.3K]

Answer:

Total heat transfer is positive

Total work transfer is positive

Explanation:

The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.

Heat transfer to a system is positive and that transferred from the system is negative.

Also, work done by a system is positive while the work done on the system is negative.

Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)

Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).

5 0

3 years ago

A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po

saul85 [17]

The location of the point F that partitions a line segment from D to E ( $\overline{DE}$ ), that goes from negative 4 to positive 5, into a 5:6 ratio is fifteen halves (option 4).

We need to calculate the segment of the line DE to find the location of point F.

Since point D is located at negative -4 and point E is at positive 5, we have:

$\overline{DE} = E - D = 5 - (-4) = 9$

Hence, the segment of the line DE ( $\overline{DE}$ ) is 9.

Knowing that point F partitions the line segment from D to E ( $\overline{DE}$ ) into a 5:6 ratio, its location would be:

$F = \frac{5}{6}\overline{DE} = \frac{5}{6}9 = 5*\frac{3}{2} = \frac{15}{2}$

Therefore, the location of point F is fifteen halves (option 4).

Learn more about segments here:

brainly.com/question/24472171?referrer=searchResults

brainly.com/question/13270900?referrer=searchResults

I hope it helps you!

5 0

3 years ago

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A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

Kobotan [32]

Answer:

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=5-9.81\times t\\\Rightarrow \frac{-5}{-9.81}=t\\\Rightarrow t=0.51 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=5\times 0.51+\frac{1}{2}\times -9.81\times 0.51^2\\\Rightarrow s=1.27\ m$

So, the stone would travel 1.27 m up

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.27+0^2}\\\Rightarrow v=5\ m/s$

Velocity as the rock passes through the original point is 5 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s$

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s$

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s$

Speed of the ball at 15 m above the ground is 17.8542 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s$

$v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s$

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

$745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229$

Coefficient of sliding friction is 0.229

8 0

3 years ago