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DanielleElmas [232]
2 years ago
13

Based on molecular orbital theory, the only molecule in the list below that has unpaired electrons is ________.

Physics
1 answer:
xenn [34]2 years ago
6 0
The answer is Radicals.
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Physics question i appreciate your help please
Over [174]

B. Velocity

Velocity is a vector quantity which means it is direction aware.

7 0
3 years ago
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Ariana is heating a beaker of water. She notices that the water at the bottom
RSB [31]
B will be the answer ( Radiation)
6 0
4 years ago
A parking lot is going to be 40 m wide and 100 m long. Which dimensions
dmitriy555 [2]
Answer: A

Explanation: You need to find the dimensions of the scale model of the lot that can be multiplied equally to get the original length. Basically 40 m and 100 m need to be divided by the same number to make the scale model evenly, and right off the bat you see that 10 and 25 could be multiplied by 4 and equal the original numbers. Hope this helps! Also sorry if it was kinda hard to understand lol, when i explain things its hard to get out of my own head ;P
8 0
3 years ago
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A 73-kg Norwegian olympian ski champion is going down a hill sloped at 39 ◦ . The coefficient of kinetic friction between the sk
bazaltina [42]

Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

= 367 N

impulse = net force x time = change in momentum .

= 367 x 5 = 1835 kg m /s

velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

net force becomes zero

mg ( sin39 - μ cos39 ) = 0

μ = tan39

= .81

c )

net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s

so he will have speed of 25.13 m /s after 5 s .

5 0
3 years ago
A 7 kg object attached to a horizontal string moves with constant speed in a circle on a frictionless horizontal surface. The ki
Marizza181 [45]

Answer:

Explanation:

Given

mass of object m=7 kg

kinetic Energy k=36\ J

Tension in string T=326\ N

mass is moving in a horizontal circle so tension is providing the centripetal acceleration

therefore T=\frac{mv^2}{r}----1

where r=radius of circle

kinetic energy of particle k=\frac{1}{2}mv^2----2

divide 1 and 2 we get

\frac{T}{k}=\frac{\frac{mv^2}{r}}{\frac{1}{2}mv^2}

\frac{T}{k}=\frac{2}{r}

r=\frac{2k}{T}=\frac{2\times 36}{326}

r=0.2208\ m

   

8 0
3 years ago
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