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Veronika [31]
3 years ago
10

The amount of gas that a helicopter uses is directly proportional to the number of hours spent flying. the helicopter flies for

44 hours and uses 2828 gallons of fuel. find the number of gallons of fuel that the helicopter uses to fly for 55 hours.
Physics
1 answer:
igomit [66]3 years ago
5 0

Answer:

The helicopter uses 35 gallons to fly for 5 hours.

Explanation:

The amount of gas that a helicopter uses for flying varies directly proportional to the number of hours spent flying.

g ∝ T

where g represents amount of gas and T time of flight.

Then,

\therefore \frac{g_1}{g_2}=\frac{T_1}{T_2}

The helicopter files 4 hours and uses 28 gallons of fuel.

Here, g₁= 28 gallons, T₁=4 hours

g₂=?, T₂=5 hours.

\therefore \frac{g_1}{g_2}=\frac{T_1}{T_2}

\Rightarrow \frac{28}{g_2}=\frac{4}{5}

⇒28×5= g₂×4

⇒ g₂×4=28×5

\Rightarrow g_2=\frac{28\times 5}{4}

\Rightarrow g_2=35 gallons

The helicopter uses 35 gallons to fly for 5 hours.

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Write the equation expressing the relationship "y varies directly as x." use k as the constant of proportionality.
Alexeev081 [22]

The equation expressing the statement "y varies directly as x"  is y=kx.

Explanation

The statement that y varies directly as x is analogous to saying that  the ratio of y to x is constant. In other words, when x increases, y likewise increases and that when x decreases, y decreases proportionally.

Mathematically, we express the relationship that the ration of y is to x is constant is expressed as; \frac{y}{x} =k where k is the constant of proportionality.

We can then solve the relationship for y to determine the correct form of the relationship as shown below,

\frac{y}{x} =k\\\rightarrow y=kx


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A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
1 year ago
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