In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Answer:
v=115 m/s
or
v=414 km/h
Explanation:
Given data
To find
Terminal velocity (in meters per second and kilometers per hour)
Solution
At terminal speed the weight equal the drag force
For speed in km/h(kilometers per hour)
To convert m/s to km/h you need to multiply the speed value by 3.6
To find the surface area of a single cube we first nees to take the cube root of 8cm3 which is 2.
Now we know that the length of each side is 2 and we can find the area of one side by doing 2x2 which is 4.
To find the total surface area of one cube we do 4 times 6 side giving us a total of 24cm2.
To find the total surface area of the 8 individual cubes, we multiply 24cm2 by 8 to give us a total of 192cm2.
Now to find the total surface area of the one large cube, we know that each side of one of the small cubes is 4cm2 and the large cube is set up so that there are two levels of four cubes right on top of each other. So, the total area of each side of the large cube is 4cm2 times 4 which gives us 16cm2.
Then we multiply 16cm2 by 6 sides to give us a total surface area of 96cm2.
The ratio of the surface area of the single large cube comapred to the total surface area of the single cubes is 96:192
We can further simplify this ratio:
96:192
48:96
24:48
12:24
6:12
3:6
1:2
Here Change in Kinetic Energy
= Work Done by Friction
Therefore, substituting the
given values to the equation, we get
0.5 * m * (vFinal^2 -
vInitial^2) = µ m g * d
Therefore
0.5*( 5.90^2 - Vfinal^2 ) =
0.100*9.8*2.10
Therefore
vfinal = 5.54 m/sec
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