Answer:
ice forming and then melting back into water
I think D x=vxt because it's equation finding change of x (displacement) and using time
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) =
and
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
So on
= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
Reduce friction because friction just makes things harder
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