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VLD [36.1K]
3 years ago
6

A proton (mass = 1.67 10–27 kg, charge = 1.60 10–19 C) moves from point A to point B under the influence of an electrostatic for

ce only. At point A the proton moves with a speed of 50 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference .A proton (mass = 1.67 10–27 kg, charge = 1.60 10–19 C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference .
Physics
1 answer:
Elena L [17]3 years ago
3 0

Answer:

20.353125 V

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

q = Charge of proton = 1.6\times 10^{-19}\ C

v_A = Velocity of proton at A = 50 km/s

v_B = Velocity of proton at B = 80 km/s

We have the relation by balancing the energy

\dfrac{1}{2}m(v_B^2-v_A^2)=q(V_B-V_A)\\\Rightarrow V_B-V_A=\dfrac{1}{2q}m(v_B^2-v_A^2)\\\Rightarrow V_B-V_A=\dfrac{1}{2\times 1.6\times 10^{-19}}\times 1.67\times 10^{-27}(80000^2-50000^2)\\\Rightarrow V_B-V_A=20.353125\ V

The potential difference is 20.353125 V

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A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert
Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, Vi and Ui are its initial vertical and horizontal components

R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi   = 10*sin(30) = 5 m/s\\Ui  = 10*cos(30) = 8.66 m/s

To find:

Max Height h achieved

Calculation:

1) Using the 3^{rd} equation of motion, we know

2*a*s = Vf^{2}  - Vi^{2}

2) In terms of gravity g height h and  the vertical component of Velocity Vf , Vi.

3) As Vf = 0 as at maximum height the vertical component of velocity is zero maximum height achieved

2*g*h = Vf^{2}  -Vi^{2}

putting values

4) h = 0.784 m/s

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

3 0
3 years ago
A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wh
maksim [4K]

Answer:16.234^{\circ}C

Explanation:

Given

Mass of object (m)=6 kg

falling height(h)=10 m

mass of water(m_w)=600 gm

temperature of water =15

specific heat of water =4.186 j/g-^{\circ}C

Let T be the Final Temperature of water

Here Object Potential Energy is converted into Heat energy which will be absorbed by water

Potential Energy(P.E.)=mgh=6\times 9.81\times 10=588.6 J

Heat supplied=m_wc(\Delta T)

H.E.=600\times 4.186\times (T-16)

588.6=2511.6\times (T-16)

T-16=0.234

T=16.234^{\circ}C

This is not an efficient way of heating water as there is only0.234^{\circ}Cincrease in temperature.

5 0
3 years ago
8. Contrast. The energy that can be released during a nuclear fission reaction with the energy that can be released during a nuc
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8) the energy released by fusion is generally 3 to 4 times larger than with fission.  Fission has very few by-products but fusion releases large amounts of radioactive particles because it starts with large nuclei.
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beta particles are high energy electrons or positrons.  They travel further due to their small size but can be stopped by a thin barrier of plastic or wood.
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7 0
2 years ago
Can anyone help me please​
ArbitrLikvidat [17]

Answer:

Gravity.

Rocket ships.

Ball.

Basketball.

Explanation:

Gravity has to do a lot with air. It puts the planets in there area.

Rocket Ship has to do a lot with air. If i'm right, they calculate the area, weather, about the air.

A ball gets throwed in the air, which gravity comes into place.

Basketball is also a similar example to a ball.

7 0
2 years ago
A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

V_{r} = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0
2 years ago
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