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timurjin [86]
3 years ago
5

An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elect

ric field. The particle then begins to move. Find the speed of the alpha particle after it has moved through a potential difference of −3.45×10^−3 V. The charge and the mass of an alpha particle are qα = 3.20×10^−19 C and mα = 6.68×10^−27 kg , respectively.
Mechanical energy is conserved in the presence of which of the following types of forces?

Select all that apply.
A- electrostatic
B- frictional
C- magnetic
D- gravitational

Part B- Which of the following quantities are unknown?
A- the initial speed of the alpha particle
B- the value of the electric potential at the initial position of the alpha particle
C- the value of the electric potential at the final position of the alpha particle
D- the final speed of the alpha particle
E- the charge of the alpha particle
F- the difference in potential between the initial and final positions of the alpha particle
G- the mass of the alpha particle

Part C- Use conservation of energy (Ki+qVi=Kf+qVf) to solve for the final kinetic energy. Then use this value to solve for the final velocity of the alpha particle.
Physics
1 answer:
alexgriva [62]3 years ago
8 0

Answer:

Explanation:

Kinetic energy gained by alpha particle

= charge x potential difference

1/2 mv² = 3.2 x 10⁻¹⁹ x 3.45 x 10⁻³

.5 x 6.68 x 10⁻²⁷ V² = 11.04 x 10⁻²²

V² = 3.3  x 10⁵

V = 5.74 x 100

= 574 m / s

Mechanical energy is conserved in respect of A , C and D .

Part B

B , C, are unknown .

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A ray of red light in air is incident at an angle of 30. on a
Vlad [161]

Answer:

20 degrees.

Explanation:

From Snell’s law of refraction:

sinθ1•n1 = sinθ2•n2

where θ1 is the incidence angle, θ2 is the refraction angle, n1 is the refraction index of light in medium1, and n2 is the refraction index for virgin olive oil. The incidence angle of the red light is θ1 = 30 degrees.

The red light is in air as medium1, so n1 (air) = 1.00029

So, to find θ2, the refracted angle:

sinθ1•1.00029 = sinθ2•1.464

sin(30)•1.00029 / 1.464 = sinθ2

0.5•1.00029 / 1.464 = sinθ2

sinθ2 = 0.3416291

θ2 = arcsin(0.3416291)

θ2 = 19.976 degrees

To the nearest degree,

θ2 = 20 degrees.

8 0
2 years ago
What is the speed shown in the graph? which number do i start on?
den301095 [7]

Answer:

4 m/s

Explanation:

speed = distance/time

speed= 20/5 = 4

similarly for all no. the answer is constant,i.e. 4

3 0
3 years ago
Un the way to the moon, the Apollo astro-
kherson [118]

Answer:

Distance =  345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2}  } \\a=3.33*10^{19} [m/s^2]

6 0
3 years ago
Is there carbon atoms in the compound Ca3N2​
katen-ka-za [31]

No, there are not any carbon atoms inside this compound.

The compound is Ca_3N_2. This means there are 3 "Ca" atoms and 2 "N" atoms.

  • Ca is calcium
  • N is nitrogen

Thus, none of the elements in this compound are carbon, meaning there are no carbon atoms. Let me know if you need any clarifications, thanks!

~ Padoru

3 0
3 years ago
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What types of landforms can water form?
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Lakes and rivers can be formed landforms from water
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