1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vlabodo [156]
3 years ago
10

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

altitude of 10 km. Assume g = 9.8 m/s' throughout the process What is the aircraft's change in potential energy? What is the aircraft's change in kinetic energy? a. b. Answers: About 1000 and 150 MJ, respectively
Physics
1 answer:
Natasha_Volkova [10]3 years ago
7 0

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

\Delta P.E=mg(h_{2}-h_{1})

\Delta P.E=10000\times9.8\times(10000-0)

\Delta P.E=10000\times9.8\times10000

\Delta P.E=980000000\ J

\Delta P.E=980\ MJ

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)

\Delta K.E=148298642\ J

\Delta K.E=148.3\ MJ

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

You might be interested in
You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

5 0
3 years ago
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
2 years ago
The volume of the moon is about 33% of the volume of the earth.true or false
stealth61 [152]
The moons volume is that of 2 percent of the earth.
4 0
3 years ago
Read 2 more answers
Which pair of equations can describe the path of a particle moving with an acceleration that is perpendicular to the velocity of
antoniya [11.8K]

Answer:

The question clearly describes the circular motion.

The circular motion equation is

a_{radial} = \frac{v^2}{r}

The path of the particle is circular.

Explanation:

In circular motion, the radial acceleration is always towards the center and constant in magnitude. Furthermore, the velocity of the circular motion is always tangential to the circle, that is it is always perpendicular to the radius, hence the acceleration.

6 0
2 years ago
A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the
Semenov [28]

Answer:

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, v, which is given by

v=\omega r

where \omega is the angular speed and r is the radius of the circular path.

\omega is given by

\omega = 2\pi f

where f is the frequency of revolution.

Thus

v=2\pi fr

Using values from the question,

v=2\pi\times 1.50\times0.75

<em>Note the conversion of 75 cm to 0.75 m</em>

v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07

6 0
3 years ago
Other questions:
  • Which of the following is a vector quantity?
    9·1 answer
  • A car has a mass of 1 200 kg. A very strong weightlifter attempts, unsuccessfully, to lift the car by applying an upward force o
    5·1 answer
  • When a beam of light passes at an oblique angle into a material of lower optical density, the angle of incidence is
    13·2 answers
  • What does Newton’s law of universal gravitation say about the distance between objects?
    15·1 answer
  • Projectile Motion: A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when
    13·1 answer
  • Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s
    8·1 answer
  • A chair is at rest on the floor. The chair absorbs thermal energy from the floor, and begins moving spontaneously with kinetic e
    12·1 answer
  • Part AIf the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negativ
    13·1 answer
  • 1. List a similarity between magnetic force and electrical force.
    5·1 answer
  • A force of 6.0 Newtons is applied to a block at rest on a horizontal frictionless surface over a 7.0 meter span. How much energy
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!