Well, st first we should find <span>initial momentum for the first person represented in the task which definitely must be :
</span>

And then we find the final one :

Then equate them together :
So we can get the velocity, which is

In that way, according to the main rules of <span>conservation of momentum you can easily find the solution for the second person.
Regards!</span>
Answer:
133.8 N
Explanation:
Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2
Therefore, the weight of the dog on this planet would be:
Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N
Answer:
a
Solid Wire
Stranded Wire 
b
Solid Wire
Stranded Wire
Explanation:
Considering the first question
From the question we are told that
The radius of the first wire is 
The radius of each strand is 
The current density in both wires is 
Considering the first wire
The cross-sectional area of the first wire is

= >
= >
Generally the current in the first wire is

=> 
=>
Considering the second wire wire
The cross-sectional area of the second wire is

=> 
=> 
Generally the current is

=> 
=> 
Considering question two
From the question we are told that
Resistivity is 
The length of each wire is 
Generally the resistance of the first wire is mathematically represented as
=>
=>
Generally the resistance of the first wire is mathematically represented as
=>
=>
Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
Hey how's your day going
I hope after I answer that you understand and don't just paste my answer into your assignment!!! (<- read!!!)
Answer \|/
Ice is less dense than water.
Reason why \|/
When water freezes the molecules inside completely stop moving (They still vibrate but don't change their position much). In doing so, they spread out a touch which makes it less dense than liquid water. So ice floats