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tester [92]
3 years ago
15

Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of v=0.8c

and determines its lifetime to be τA=129 ns. Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory?
Physics
1 answer:
Studentka2010 [4]3 years ago
3 0

Answer:

30.96 m

Explanation:

If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:

d = v * Δt

v = 0.8*c = 0.8 * 3e8 = 2.4e8

Δt = ζ = 129 ns = 1.29e-7 s

d = 2.4e8 * 1.29e-7 = 30.96 m

This is the distance as measured by observer A.

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It could possibly melt things
6 0
3 years ago
A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor
STatiana [176]

Answer:

The magnitude of the second charge is \rm 1.062\times 10^{-7}\ C or \rm 0.1062\ \mu C.

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge q_o at a point in the electric field of another charge q is given by the product of the amount of charge q_o and electric potential at that point due to the charge q.

U = q_o\ V.

The electric potential at that point is given by

V = \dfrac{kq}{r}.

where k is the Coulomb's constant.

Therefore,

U=q_o\ \dfrac{kq}{r}.

Now, We have given two charges q_1 = +44\ \mu C = +44\times 10^{-6}\ C and q_2, whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

U_i = \dfrac{kq_1q_2}{\infty}=0.

When the coordinates of position of the two charges are

(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).

The distance between the two charges is given by

r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.

The electric potential energy of the charges in this configuration is given by

U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.

6 0
3 years ago
A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.​
disa [49]

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

6 0
2 years ago
A 70 kg base runner begins his slide into second base while moving at a speed of 4.35 m/s. He slides so that his speed is zero j
sladkih [1.3K]

Answer

given,

Mass of the runner, M = 70 Kg

speed of the runner on the second base = 4.35 m/s

speed at the base = 0 m/s

Acceleration due to gravity,g = 9.8 m/s²

a) magnitude of mechanical energy lost

  Mechanical energy lost is equal top gain in kinetic energy

   ME_{Lost}=\dfrac{1}{2}mv^2

   ME_{Lost}=\dfrac{1}{2}\times 70\times 4.35^2

   ME_{Lost}=662.29\ J

b) Work done = Force x displacement

    W = F. x

     F = μ mg

    W = μ mg . x

Work done is equal to 662.29 J

  x=\dfrac{W}{\mu m g}

using the coefficient of the friction,μ = 0.7

  x=\dfrac{662.29}{ 0.7\times 70\times 9.8}

     x = 1.38 m

Hence, the runner will slide to 1.38 m.

3 0
2 years ago
What is the definition of force according to Newton's first law of motion?​
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Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force
8 0
3 years ago
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