Answer:
Distance will be 
Explanation:
We have given that Buffalo has a latitude of 42.9°N
And Raleigh has a latitude of 35.8°N
Radius of the earth = 3960 miles
We have to calculate the distance between given two cities
Difference in their latitudes = 
Now changing the angle in radian = 
So distance will be 
Answer:
-0.7 m/sec
Explanation:
Mass of first block = m1 =3.0 kg
Mass of second block = m2= 5.0 kg
Velocity of first block = V1= 1.2 m/s
Velocity of second block = V2 = ?
Momentum of Center of mass MVcom is sum of both blocks momentum and is given by
MVcom= m1v1+m2v2
Where
M= mass of center of mass
Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)
Putting values, we get;
0= 3×1.2+5v2
==> v2= 3.6/5= - 0.7 m/s
-ve sign indicates that block 2 is moving in opposite direction of block 1
Answer:
The current is not used up. The electrons flow through the entire circuit and this travel is the current. They flow until they are not charged anymore. That is also why the circuit must be closed or else electrons would escape not just light it up for a second then go out.
Explanation:
The third choice is correct
Answer:
The upper limit on the flow rate = 39.46 ft³/hr
Explanation:
Using Ergun Equation to calculate the pressure drop across packed bed;
we have:

where;
L = length of the bed
= viscosity
U = superficial velocity
= void fraction
dp = equivalent spherical diameter of bed material (m)
= liquid density (kg/m³)
However, since U ∝ Q and all parameters are constant ; we can write our equation to be :
ΔP = AQ + BQ²
where;
ΔP = pressure drop
Q = flow rate
Given that:
9.6 = A12 + B12²
Then
12A + 144B = 9.6 -------------- equation (1)
24A + 576B = 24.1 --------------- equation (2)
Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So
288 B = 4.9
B = 0.017014
From equation (1)
12A + 144B = 9.6
12A + 144(0.017014) = 9.6
12 A = 9.6 - 144(0.017014)

A = 0.5958
Thus;
ΔP = AQ + BQ²
Given that ΔP = 50 psi
Then
50 = 0.5958 Q + 0.017014 Q²
Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;
Q² + 35.02Q - 2938.8 = 0
Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;
Q = 39.46 ft³/hr