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2 years ago

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

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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)]

(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)]

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