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
maksim [4K]
3 years ago
15

What happens to the kinetic energy of a roller coaster as it approaches its lowest point and loses its potential energy?

Physics
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

At the top of the roller coaster, there is a lot of potential energy. When it comes to the bottom, the roller coaster loses its potential energy and gains kinetic energy as it is going very fast here.

You might be interested in
A bird sits on top of a 639 m tall tower. If it's gravitational potential energy up there is 2033 J, what is its mass?
iris [78.8K]

The mass of the bird is 0.32 kg.

<u>Explanation:</u>

Gravitational potential energy, the energy exhibited by an object at rest due to the influence of gravitational force. So the increase in distance of object from the surface of earth leads to increase in the gravitational potential energy. Thus,

       \text {Gravitational potential energy}=m \times \text { Acceleration } \times \text { Distance of bird from bottom }

So, as the gravitational potential energy is given as 2033 J and the position of bird placed on the tall tower is 639 m away from the bottom, then the mass (m) of the bird can be found as below.

       m o f \text { bird }=\frac{\text {Gravitational potential energy}}{a \times \text {Distance}}=\frac{2033}{9.8 \times 639}=\frac{2033}{6262.2}

So, finally we get the bird's mass as,

            m of bird = 0.32 kg

7 0
3 years ago
A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the
vredina [299]

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light  λ is  580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D  = 1.8 m.

Width of one fringe = \frac{\lambda\times D}{d}

Putting the values we get fringe width

= \frac{580\times10^{-9}\times1.8}{.000315}

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe  is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be  

= 4.86 mm.

3 0
3 years ago
The rate of cooling determines ....... and ......​
PilotLPTM [1.2K]

Answer:

freezing point and melting point

7 0
3 years ago
When you must give something up in order to get something else, it is called...
natali 33 [55]
Opportunity cost refers to what you have to give up to buy what you want in terms of other goods or services. When economists use the word “cost,” we usually mean opportunity cost.
6 0
3 years ago
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
Other questions:
  • A model airplane has momentum given by p=[(-0.75kg.m/s3)t2 + (3.0kg.m/s)] i + (0.25kg.m/s2)t j. Find the components Fx, Fy, and
    5·1 answer
  • Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu
    15·1 answer
  • Has an atomic number of 1
    11·2 answers
  • Which is an example of a chemical change?
    8·1 answer
  • A 4,350 kg truck traveling east has a momentum of 26,100kg m/s what's the trucks velocity
    14·1 answer
  • A kayak took 5 hours to finish its trip on a river. If it traveled at an
    6·1 answer
  • water is know to boil at 100°C.A student boiled water and realised it's boiling point was 101°C.State two possible reasons ​
    5·1 answer
  • What is the speed of an object at a given moment
    9·1 answer
  • What is the magnitude of velocity for a 2,000 kg car possessing 3,000 kg(*)m/s of momentum?
    8·1 answer
  • The two forces to the right are friction which one would be the force of kinetic friction?
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!