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
RideAnS [48]
3 years ago
10

A stone is thrown vertically up. What kind of energy did the stone have initially? What happens to this energy as the stone asce

nds?
Physics
1 answer:
Alex3 years ago
7 0

Answer:

kinetic energy at first

Explanation:

kinetic turns to potential as it gains height

You might be interested in
A 2.0-kg block is on a perfectly smooth (frictionless) ramp that makes an angle of 30^\circ30 ​∘ ​​ with the horizontal. What is
Nastasia [14]

Answer:

Explanation:

Since the surface is frictionless therefore there will be no friction force on block but there will be weight of block which we can divide in to two components i.e. mgcosθ &mgsinθ which is perpendicular and parallel to the surface respectively.

In response to mgcosθ ramp will apply a normal force to the block which will be of equal magnitude to that of mgcosθ.

Therefore Ramp will apply a Force of mgcosθ on block where m is the mass of block.

7 0
2 years ago
Any five chapter 9 institutions found in the constitution of south africa
navik [9.2K]
Five institutions from the chapter 9 found in the constitution of south Africa are : 
- The Public Protector
- The south African Human right commission
- The commission of Gender Equality
- The Auditor- General
- The Independent Electoral Commission

Hope this helps
7 0
3 years ago
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
2 years ago
Read 2 more answers
Your friend tells you “I no the moon does not rotate because we always see the same side.” do you agree or disagree with your fr
nydimaria [60]

Answer:

no the moon does not rotate it only goes in circle just like the sun so I disagree with your friend

3 0
3 years ago
Which statement best describes the effect of the magnet on the block of
barxatty [35]

The magnet (south pole of the magnet) has magnetized the right side of the block.

<h3>Direction of electric field in the magnetic material</h3>

The direction of electric field of the atom of the magnetic material is unpolarized.

From the diagram in the image, the right hand side of the magnetic material is being attracted to south pole of the magnet.

Thus, we can conclude that, the magnet has magnetized the right side of the block.

Learn more about magnetic material here: brainly.com/question/22074447

#SPJ1

7 0
2 years ago
Other questions:
  • A fairgrounds ride spins its occupants inside a flying saucer-shaped container. if the horizontal circular path the riders follo
    12·1 answer
  • why people change their way of being They can be temporary or permanent changes. Other reasons why people change their attitudes
    9·1 answer
  • A 13,500 kg railroad freight car travels on a level track at a speed of 4.5 m/s. It collided and coupled with a 25,000 kg second
    8·1 answer
  • Name 10 transition metals
    9·1 answer
  • A block of mass m=3.0 kg is pushed down a 50° incline surface with a force of magnitude F=40 N.
    15·1 answer
  • Two horizontal forces, 230 N and 120 N, are exerted in opposite direction on a crate. What is the horizontal acceleration of the
    9·1 answer
  • What is the geocentric angle of the earthquake focus?​
    14·1 answer
  • ¿Qué pueden decirnos las capas de hielo del Ártico sobre nuestros cambios atmosféricos?
    15·1 answer
  • A book moving across a desk has a net force equaling 60 N, to the right.
    14·1 answer
  • What is generally TRUE about diagnosing psychological disorders?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!