Mass- Mass is measured in kilograms (kg).
Weight- Weight is measured in Newton’s.
Answer:
A scalar is a quantity that is fully described by a magnitude only. It is described by just a single number. Some examples of scalar quantities include speed, volume, mass, temperature, power, energy, and time.
Examples of scalar quantity are:
Distance.
Speed.
Mass.
Temperature.
Energy.
Work.
Volume.
Area.
Explanation:
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Answer:
It is the carrier of genetic information.
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
Answer:0.58 m
Explanation:
The initial velocity of the ball is u = 2.0 m/s
The height of the table is, h = 1.0 m
The ball falls in vertical direction under acceleration due to gravity.
Time taken for ball to hit the floor:
h= ut + 0.5gt² ( from the equation of motion)
1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²
Solving this for t,
t = 0.29 s ( we have neglected the negative value of t)
In the same time, the ball would cover a horizontal distance of :
s = u t
⇒s = 2.0 m/s×0.29 s = 0.58 m
Thus, the landing spot is 0.58 m away from the table.