Using the picture provided the forces are added together, because they are putting force on an object the same direction, thus the forces are added.
Answer:
Δv = 12 m/s, but we are not given the direction, so there are really an infinite number of potential solutions.
Maximum initial speed is 40.6 m/s
Minimum initial speed is 16.6 m/s
Explanation:
Assume this is a NET impulse so we can ignore friction.
An impulse results in a change of momentum
The impulse applied was
p = Ft = 1400(6.0) = 8400 N•s
p = mΔv
Δv = 8400 / 700 = 12 m/s
If the impulse was applied in the direction the car was already moving, the initial velocity was
vi = 28.6 - 12 = 16.6 m/s
if the impulse was applied in the direction opposite of the original velocity, the initial velocity was
vi = 28.6 + 12 = 40.6 m/s
Other angles of Net force would result in various initial velocities.
Choice-C is a correct statement.
-- The area under a velocity/time graph, between two points in time, is the difference in displacement during that period of time.
-- The area under a speed/time graph, between two points in time, is the distance covered during that period of time.
The options are;
a. V2 equals 2V1.
b. V2 equals (V1)/2.
c. V2 equals V1.
d. V2 equals (V1)/4.
e. V2 equals 4V1.
Answer:
Option A: V2 equals 2V1
Explanation:
Since the flow is steady, then we can say;
mass flow rate at input = mass flow rate at output.
Formula for mass flow rate is;
m' = ρVA
Thus;
At input;
m'1 = ρ1•V1•A1
At output;
m'2 = ρ2•V2•A2
So, m'1 = m'2
Now, we are told that the density of the fluid decreases to half its initial value.
Thus; ρ2 = (ρ1)/2
Since m'1 = m'2, then;
ρ1•V1•A1 = (ρ1)/2•V2•A2
Now, the pipe is uniform and thus the cross section doesn't change. Thus;
A1 = A2
We now have;
ρ1•V1•A1 = (ρ1)/2•V2•A1
A1 and ρ1 will cancel out to give;
V1 = (V2)/2
Thus, V2 = 2V1
