Answer:
The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.
Explanation:
Data
F = 7 N
F = ? if the masses is quartered
Formula
Process
Normal conditions F = Km₁m₂/r² = 7
When masses quartered F = K(m₁/4)(m₂/4)/r² = ?
F = K(m₁m₂/16)/r²
F = K(m₁m₂/16r² = 7/16 = 0.4375 N
Answer:
130.5
Explanation:
According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.
The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:
16 | 8 5 6
This means that the point for the students are 168, 165 and 166. Three students, three points.
If you watch the stemplot, the points for the students are:
116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.
The median can be calculated using the mean between the two values in the middle of the sequence.
In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:
Median = 128 + 133 / 2 = 130.5
Answer:
Explanation:
1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is
100kg/h*0.5h = 50kg
2. the momentum does not conserve because the drag force of water makes that the boat loses velocity
3. If we assume that the force of the boat before the raining is
where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts
And if we take the net force as
where we take v=1m/s because we are taking into account tha velocity just after the rain stars.
I hope this is useful for you
regards