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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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AURORKA [14]

Answer:

0.084 kg

Explanation:

I = 0.80 N-s (East wards) = 0.80 i N-s

u = 3.8 m/s = - 3.8 i m/s

v = 5.7 m/s = 5.7 i m/s

Let m be the mass of bat.

I = m (v - u)

0.8 i = m ( 5.7 i + 3.8 i)

0.8 i = m x 9.5 i

m = 0.084 kg

8 0
3 years ago
PLEASE help me!!! ASAP!! Ill mark you as brainliest!!!!
MakcuM [25]

Answer:

B

Explanation:

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8 0
2 years ago
A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to
sineoko [7]
The temperature of the water getting colder would cause the liquid in the thermometer to drop due to less heat being transferred from the water to the liquid, so the liquid molecules are closer than when they have high energy.
3 0
3 years ago
Read 2 more answers
What resistance would produce a current of 200A with a potiential difference of 200V?
viktelen [127]
Aw, I hate physics, is this on Apex?

Resistance can be calculated with the information given in the question.
Equation for Resistance: R = V/I
V (voltage) = 200 Volts
I (current) = 200 Amps

So 200 divided by 200 = freaking 1

Answer: R = 1 (ohms)

Hope this Helps!

5 0
3 years ago
The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting
White raven [17]

Answer:

\frac{\dot Q}{A} =20.129\ W.m^{-2}

T_1=27.58\ ^{\circ}C & T_2=2.41875\ ^{\circ}C

Explanation:

Given:

  • interior temperature of box, T_i=30^{\circ}C
  • height of the walls of box, h=3\ m
  • thickness of each layer of bi-layered plywood, x_p=1.25\ cm=0.0125\ m
  • thermal conductivity of plywood, k_p=0.104\ W.m^{-1}.K^{-1}
  • thickness of sandwiched Styrofoam, x_s=5\ cm=0.05\ m
  • thermal conductivity of Styrofoam, k_s=0.04\ W.m^{-1}.K^{-1}
  • exterior temperature, T_o=0^{\circ}C

<u>From the Fourier's law of conduction:</u>

\dot Q=\frac{dT}{(\frac{x}{kA}) }

\dot Q=\frac{dT}{R_{th} } ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

R_{th}=R_p+R_s+R_p

R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}

R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})

R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})

R_{th}=\frac{1.4904}{A} .....................(2)

Putting the value from (2) into (1):

\dot Q=\frac{30-0}{\frac{1.4904}{A} }

\dot Q=\frac{30\ A}{1.4904}

\frac{\dot Q}{A} =20.129\ W.m^{-2} is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}

20.129=0.104\times \frac{30-T_1}{0.0125}

T_1=27.58\ ^{\circ}C

&<u>For Styrofoam-plywood interface from inside:</u>

\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}

20.129=0.04\times \frac{27.58-T_2}{0.05}

T_2=2.41875\ ^{\circ}C

4 0
3 years ago
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