To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).
The answer is 36 kg m/s
Answer: Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form. An internal combustion engine provides a good example of the ease with which gases can be compressed.
Explanation:
Answer:
Explanation:
We know that the pressure can be calculated in the following way:
p = d·g·h
with d being the density of the water, g the gravitational acceleration and h the depth.
Also d of the water = 1000 kg/m^3 circa and g = 9.8 m/s^2 circa
117,500 Pa = 1000kg/m³ · 9.8m/s² · h
Therefore h = 11,9 m
Answer:
Electric field E = kQ/r^2
Distance between charges = 6.30 - (-4.40) = 10.70m
Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.
Field from q1 at P = k(-9.50x^10^-6) / d^2
Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2
These fields are in opposite directions and are equal magnitudes if the resultant field = 0
k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2
9.50 / d^2 =8.40 / (10.70-d)^2
d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131
d/(10.70-d) = sqrt(1.1331) = 1.063
d = 1.063 ((10.70-d)
= 10.63 - 1.063d
2.063d = 10.63
d = 5.15m
The y coordinate where field is zero is 6.30 - 5.15 = 1.15m
Explanation:
To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:
We know that the flow rate is equivalent to the velocity of the fluid in its area, that is,
Where
V = Velocity
A = Cross-sectional Area
Our values are given as
Since there is continuity we have now that,
Therefore the speed of the water's house supply line is 0.347m/s
