Answer: The correct answer is option (C).
Explanation:
As it is given in the problem, the path of a meteor passing Earth is affected by its gravitational force and falls to Earth's surface. Another meteor of the same mass falls to Jupiter's surface due to its gravitational force.
According to Newton's law of universal gravitational, every particle attracts every other particles in the universe with the gravitational force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The Jupiter is the most massive planet in the solar system. It is also the largest planet in the solar system. The gravity of Jupiter on its surface is 2.4 times that of surface gravity of the Earth.
If a person weighs 100 pounds on the Earth then he would weigh 240 pounds on Jupiter.
Therefore, the correct answer is option (C), the meteor falls to Jupiter faster due to its greater gravitational force.
Answer:
True
Explanation:
I am not 100% on the answer for this question but i hope it was right
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
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
<h2>Answer: 10.52m</h2><h2 />
First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).
According to this, the initial velocity
has two components, because the brick was thrown at an angle
:
(1)
(2)
(3)
(4)
As this is a projectile motion, we have two principal equations related:
<h2>
In the x-axis:
</h2>
(5)
Where:
is the distance where the brick landed
is the time in seconds
If we already know
and
, we have to find the time (we will need it for the following equation):
(6)
(7)
<h2>
In the y-axis:
</h2>
(8)
Where:
is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>
is the acceleration due gravity
Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:
(9)
(10)
Multiplying by -1 each side of the equation:
>>>>This is the height of the building