Any substance that yields a hydrogen ion when placed in a water solution is called a(n) _____.
2 answers:
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Answer:
Explanation:
Considering the fact that we ave been given an angle of inclination here, we best use it! That means that the velocity of 23 m/s is actually NOT the velocity we need; I tell my students that it is a "blanket" velocity but is not accurate in either the x or the y dimension of parabolic motion. In order to find the actual velocity in the dimension in which we are working, which is the y-dimension, we use the formula:
and filling in:
which gives us an upwards velocity of 9.7 m/s. So here's what we have to work with in its entirety:
a = -9.8 m/s/s
t = 2.8 seconds
Δx = ?? m
The one-dimensional motion equation that utilizes all of these variables is
Δx = and filling in:
Δx = I am going to do the math according to the correct rules of significant digits, so to the left of the + sign and to 2 sig fig, we have
Δx = 27 + and then to the right of the + sign and to 2 significant digits we have
Δx = 27 - 38 so
Δx = -11 meters. Now, we all know that distance is not a negative value, but what this negative number tells us is that the ball fell 11 meters BELOW the point from which it was kicked, which is the same thing as being kicked from a building that is 11 meters high.
Answer:
Speed of ball A after collision is 3.7 m/s
Speed of ball B after collision is 2 m/s
Direction of ball A after collision is towards positive x axis
Total momentum after collision is m×4·21 kgm/s
Total kinetic energy after collision is m×8·85 J
Explanation:
<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>Let the mass of each ball be m kg
v be the velocity of ball A along positive x axis
v be the velocity of ball A along positive y axis
u be the velocity of ball B along positive y axis
Conservation of momentum along x axis
m×3·7 = m× v
∴ v = 3.7 m/s along positive x axis
Conservation of momentum along y axis
m×2 = m×u + m× v
2 = u + v → equation 1
∴ relative velocity of approach = relative velocity of separation
-2 = v - u → equation 2
By adding both equations 1 and 2 we get
v = 0
∴ u = 2 m/s along positive y axis
Kinetic energy before collision and after collision remains constant because it is an elastic collision
Kinetic energy = (m×2² + m×3·7²)÷2
= 8·85×m J
Total momentum = m×√(2² + 3·7²)
= m× 4·21 kgm/s
Answer:
0.00354 (N)
Explanation:
Convert to metric system:
Formula for gravitational force:
where s is the distance between 2 bodies masses m and M
Substitute the number to the formula above and since the 2 forces are acting in opposite direction, the total net gravitational force on the mass of origin be: