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Nuetrik [128]
3 years ago
15

What can i say about the magnetic field? (in image)

Physics
2 answers:
Svetlanka [38]3 years ago
7 0
I think the intensity is very low so director must be in support of the wire
gizmo_the_mogwai [7]3 years ago
3 0
I think it’s, its direction would reinforce the original movement
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a feather is dropped on the moon from a height of 1.40meters. the acceleration of gravity on the moon is 1.67m/s^2. determine th
kykrilka [37]

Answer:

1min since there is no gravity on the moon so it will take time to drop on the moon.

Explanation:

6 0
3 years ago
The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug
9966 [12]

Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

6 0
4 years ago
Read 2 more answers
E=<br> (500.0lm)<br> 4 (100)
drek231 [11]

Answer:

didn't understand your question

7 0
3 years ago
A. How far does a 100-newton force have to move to do 1,000 joules
Aloiza [94]

Work done by a force is given as the product of force and the distance moved by the force.

<h3>What is work done?</h3>

Work done is the product of force and the distance moved by the the force.

  • Work done = Force × distance

Thus, distance required by the 100 N force is given as:

  • Distance = work done/force

Distance = 1000/100 = 10 m

Distance to be moved is 10 m.

Force applied = work done/ distance

Force applied by the hoist = 500/2

Force applied by the hoist = 250 N

Distance moved in one push-up = 25 cm = 0.25 m

Work done by the athlete after one push-up = 250 × 0.25 m

Work done by the athlete = 62.5 J

Distance moved by the force = 0 m

Work done = 500 × 0 = 0 N

Therefore, for work to be done, force has to move a distance.

Learn more about work done at: brainly.com/question/25573309

5 0
2 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
3 years ago
Read 2 more answers
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