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
Leviafan [203]
2 years ago
12

How many degrees are there between the direction of motion and the force of

Physics
2 answers:
Aleks04 [339]2 years ago
5 0

Answer:

180°

Explanation:

Friction, if it exists, ALWAYS opposes motion or attempted motion.

Anna71 [15]2 years ago
3 0
C. 180
Friction yes no maybe
You might be interested in
The brachialis attaches to the forearm .035 m from the elbow at an angle of 32 deg. If the brachialis produces 750 N of force, w
jek_recluse [69]

Answer:

XY=636N

Explanation:

From the question we are told that:

Distance d=0.35m

Angle \theta=32\textdegree

Force F=750N

Generally the equation for magnitude of the stabilizing component of the brachialis force is mathematically given by

 XY=Fcos\theta

 XY=750cos 32\textdegree

 XY=636N

4 0
3 years ago
Is Li Chen abiotic or biotic?
kherson [118]

Answer:

abiotic

Explanation:

i think but dont take my word for it

8 0
3 years ago
Read 2 more answers
A. What is the RMS speed of Helium atoms when the temperature of the Helium gas is 343.0 K? (Possibly useful constants: the atom
kkurt [141]

Answer:

(a) 1462.38 m/s

(b) 2068.13 m/s

Explanation:

(a)

The Kinetic energy of the atom can be given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 10⁻²³ J/k

K.E = Kinetic Energy of atoms = 343 K

T = absolute temperature of atoms

The K.E is also given as:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

v² = 3KT/m

v = √[3KT/m]

where,

m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg

v = RMS Speed of Helium Atoms = ?

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 1462.38 m/s</u>

(b)

For double temperature:

T = 2 x 343 K = 686 K

all other data remains same:

v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 2068.13 m/s</u>

8 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
What Do You Get When You Multiply An Object's mass times the acceleration?
emmasim [6.3K]

You get the net force acting on it ... the sum of the strengths and directions
of all the individual forces there may be. 


7 0
3 years ago
Other questions:
  • 4. Inflection best relates to A. vocabulary. B. volume. C. tone. D. pace.
    12·1 answer
  • Why is it important to have a balance of gases in the atmosphere?
    5·1 answer
  • HELPPPPP ASAPPP!!!!!!!
    9·1 answer
  • Reto: Aníbal desea construir una carpa de base hexagonal de 1,40 m de altura. El vértice de la base es de 0,80m y la longitud de
    6·1 answer
  • Which of the following is not a common property of basesp
    7·1 answer
  • A water pump with a power of 3.4 × 102 watts lifts water at the rate of 7.5 × 10-2 meters/second from a water tank. What is the
    8·1 answer
  • Graph the following data tables on different graphs.
    13·1 answer
  • One of the characteristics of the planet mars that makes it very different from earth is that it does not have a large magnetic
    7·2 answers
  • The particle in the atom with a negative charge is the ______<br> Answer here
    13·2 answers
  • What happens to your breathing rate when you.a) exercise.b) go to sleep
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!