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:
Option (c) : 20°C
Explanation:
T(final) = 500* 10 + 100*70/600 = 20°C
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
Answer:
Rutherford and atomic model are correctly matched.
2 J / 4 C = 1/2 joule per coulomb.
That's 1/2 volt.
