Answer:
The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Explanation:
To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.
By definition it is known that the conservation of the moment is given by:

Our values are given by,

As the skater 1 run in x direction, there is not component in Y direction. Then,
Skate 1:


Skate 2:


Then, if we applying the formula in X direction:
m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}
75*5.45-75*1.41=(75+75)v_{fx}
Re-arrange and solving for v_{fx}
v_{fx}=\frac{4.04}{2}
v_{fx}=2.02m/s
Now applying the formula in Y direction:




Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Answer:
4 A
Explanation:
We are given that

I=12 A
We have to find the current flowing through each resistor.
We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.
Formula :

Using the formula



Substitute the values



Hence, current flows through any one of the resistors is 4 A.
To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

Where,
m = Mass
v = Velocity
Replacing we have that the Total Kinetic Energy is



On the other hand the required Energy to heat up t melting point is


Where,
m = Mass
Specific Heat
Change at temperature
Latent heat of fussion
Heat required to heat up to melting point,




The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.
Each BB 0.349 grams and 5.386 grains. 349 mg & 5.386 grains each steel BBs.
Electrical energy.................