If the balloon can barely lift an additional 3500 N of passengers, breakfast, and champagne when the outside air density is 1.23
1 answer:
The complete question is :
A hot-air balloon has a volume of 2100 m3 . The balloon fabric (the envelope) weighs 860 N . The basket with gear and full propane tanks weighs 1300 N .
If the balloon can barely lift an additional 3400 N of passengers, breakfast, and champagne when the outside air density is 1.23kg/m3, what is the average density of the heated gases in the envelope?
Solution :
Given volume of the hot air balloon
The balloon fabric weights = 860 N
The weight of the basket with the gear and propane tank = 1300 N
Density of outside air
∴ Total pay load = Weight of the air displaced - weight of gas inside the balloon
Total pay load = 860 + 1300 + 3400
= 5560 N
Mass = density x volume
Weight
Weight = volume x density
Weight of the displaced air = 2100 x 1.23 x 9.8
= 25313 N
Weight of the gas inside the balloon = density x volume
= density x 9.8 x 2100
= density x 20580 N
Therefore substituting the values, we get
⇒ 25313 - (density x 20580) = 5560
⇒ density
So the density of the heated gas
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The question is incomplete. Here is the complete question.
Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed
. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed
, which is greater than
.
Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.
Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.
Answer: Part A:
Part B:
Explanation: First, let's write an equation of motion for each car.
Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:
where
is initial position
v is velocity
t is time
Car A started the race at a distance. So at t = 0, initial position is .
The equation will be:
Car B started at the starting line. So, its equation is
Part A: When they meet, both car are at "the same position":
Car B meet with Car A after units of time.
Part B: With the meeting time, we can determine the position they will be:
Since Car B started at the starting line, the distance Car B will be when it passes Car A is units of distance.
Answer:
The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N
Explanation:
The question relates to the force of gravity experienced between two bodies
The given parameters are;
The mass of Big Ben, M₁ = 1 × 10⁸ kg
The mass of the Empire State Building, M₂ = 1 × 10⁹ kg
The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters
By Newton's Law of gravitation, we have;
Where;
F = The force exerted by Big Ben on the Empire State Building and vice versa
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
M₁, M₂, and r are the given parameters
By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;
The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.