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ivanzaharov [21]
3 years ago
11

A plane is landing at an airport. The plane has a massive amount of kinetic energy due to it's motion. When the plane lands, it

activates its brakes, causing it to slow down and eventually stop. The kinetic energy that the plane had is now gone. In keeping with the Law of Conservation of Energy, which of the following is the most likely explanation of what happened to that energy?
A. The brakes used a coil system to convert the kinetic energy into potential energy stored in the brakes
B. The kinetic energy of the plane gets pushed into the air in front of it due to drag forces, causing the air to move, thus transferring the kinetic energy from the plane to the air
C. The brakes create friction, which transformed the kinetic energy to heat energy that was dissipated to the surroundings
D. The kinetic energy is being converted into sound energy by the loud engines of the plane
Physics
1 answer:
marshall27 [118]3 years ago
8 0

Answer:

A. The brakes used a coil system to convert the kinetic energy into potential energy stored in the brakes

Explanation:

Based on the law of conservation of energy, the brakes used a coil system to convert the kinetic energy into potential energy stored in the brakes.

The law of conservation of energy states that energy is neither created nor destroyed in a system but it is transformed from one form to another.

As the airplane slows down, the kinetic energy which is presented in the motion of the plane is gradually converted to potential energy.

The potential energy is the energy due to the position of a body.

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6 0
3 years ago
An ideal gas initially at 4.00atm and 350 K is permitted
Nuetrik [128]

Explanation:

It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is P_{2} and final temperature is T_{2}.

(a)   We know that for a monoatomic gas, value of \gamma is \frac{5}{3}[/tex].

And, in case of adiabatic process,

                PV^{\gamma} = constant              

also,         PV = nRT

So, here    T_{1} = 350 K,    V_{1} = V,  and   V_{2} = 1.5 V

Hence,      \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}

         \frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{5}{3} -1}

          T_{2} = 267 K

Also,   P_{1} = 4.0 atm,   V_{1} = V,  and   V_{2} = 1.5 V

        \frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}

        \frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{5}{3}}

            P_{2} = 2.04 atm

Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.

(b) For diatomic gas, value of \gamma is \frac{7}{5}[/tex].

As,        PV^{\gamma} = constant              

also,         PV = nRT

T_{1} = 350 K,    V_{1} = V,  and   V_{2} = 1.5 V

              \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}

         \frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{7}{5} -1}

          T_{2} = 289 K

And,   P_{1} = 4.0 atm,   V_{1} = V,  and   V_{2} = 1.5 V

                \frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}

        \frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{7}{5}}

            P_{2} = 2.27 atm

Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.

6 0
4 years ago
In the oscillating spring ball system, where is the velocity of the ball the greatest?
nlexa [21]

The speed of an object in a mass-spring system is given under the function

v = \pm \sqrt{\frac{k}{m}(A^2-x^2)}

Here,

m = mass

k = Spring constant

A = Amplitude

x = Position

When the position is at the equilibrium point (x = 0), the speed will be maximum, and could even be expressed as

v_{max}= A\sqrt{\frac{x}{m}}

So the correct answer is B.

6 0
3 years ago
Explain why two acetate rods, both charged with silk repel
kenny6666 [7]
Two acetate rods, both charged with silk would repel because they are both have positively charged electrons.

Explanation: Opposite charges attract. Like charges repel.
8 0
3 years ago
A Parachutist with a camera, with descending at a speed of 12.5m/s, releases, the camera at an altitude of 64.3m. What is the ma
jonny [76]

Given :

Initial velocity, u = 12.5 m/s.

Height of camera, h = 64.3 m.

Acceleration due to gravity, g = 9.8 m/s².

To Find :

How long does it take the camera to reach the ground.

Solution :

By equation of motion :

h = ut+\dfrac{gt^2}{2}

Putting all given values, we get :

12.5t+\dfrac{9.8t^2}{2}=64.3\\\\4.9t^2+12.5t=64.3

t = 2.56  and t = −5.116.

Since, time cannot be negative.

t = 2.56 s.

Therefore, time taken is 2.56 s.

Hence, this is the required solution.

7 0
3 years ago
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