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At what speed, as a fraction of c , is a particle's total energy twice its rest energy

WINSTONCH [101]

The rest energy of a particle is
$E_0=m_0 c^2$
where $m_0$ is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
$E=mc^2 = \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }$
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
$E=2E_0$
which means:
$\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2$
$\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2$
From which we find the ratio between the speed of the particle v and the speed of light c:
$\frac{v}{c}= \sqrt{1- (\frac{1}{2})^2 } =0.87$
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.

3 0

3 years ago

Satellite A is orbiting Earth at an altitude of 500 km and Satellite B is orbiting 800 km above the surface . How does the veloc

sp2606 [1]

The velocity of Satellite A is 2% greater than velocity of satellite B.

The given parameters;

<em>Altitude of Satellite A = 500 km</em>
<em>Altitude of Satellite B = 800 km</em>

The forces acting on the Satellites are given as follows;

$F_c = \frac{mv^2}{r} \\\\F_g = \frac{GMm}{r^2} \\\\\frac{v^2}{r} = \frac{GM}{r^2}\\\\v^2 = \frac{GM}{r} \\\\v^2 r = GM\\\\v_1^2 r _1 = v_2^2 r_2\\\\v_A^2r_A = v_B^2 r_B\\\\(\frac{v_A}{v_B} )^2 = \frac{r_B}{r_A} \\\\\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A} } \\\\\frac{v_A}{v_B} = \sqrt{\frac{(800,000\ + \ 6.4 \times 10^6}{(500,000\ + \ 6.4 \times 10^6} )} \\\\\frac{v_A}{v_B} = 1.02 \\\\v_A = 1.02 \ v_B$

$v_A = v_B( 100\% \ + 2\%)\\\\v_A = 100\%v_B \ + \ \ 2\% v_B\\\\v_A = v_B \ \ + \ 2\% v_B$

Thus, the velocity of Satellite A is 2% greater than velocity of satellite B.

Learn more about velocity of satellite here: brainly.com/question/13981089

4 0

3 years ago

The KE of a body becomes 2 times of its original value then the new momentum will be more than its initial momentum by

sladkih [1.3K]

Answer:

√2

Explanation:

If the final kinetic energy is 2 times the initial kinetic energy:

KE = 2 KE₀

½ mv² = 2 (½ mv₀²)

v² = 2 v₀²

v = √2 v₀

Therefore, the ratio of the final momentum to the initial momentum is:

p / p₀

mv / (mv₀)

v / v₀

√2

7 0

3 years ago

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and at an averag

Ilya [14]

Answer:

$\dfrac{32+0.2y}{40}$

Explanation:

Total =40 miles.

y miles at x miles/hr.

40- y miles at 1.25 x miles/hr.\

We know that

Distance = time x velocity

For y miles

Lets time taken to cover y miles is t

y = x t

t=y/x -----------1

For 40- y

Lets time taken to cover 40- y is t'

40- y = 1.25 x t'

t'=(40-y)/1.25x -----------2

By adding both equation

t+t'=y/x + (40-y)/1.25x

t+t'=(32+0.2y)/x

Now lets time T taken by Marla if she will travel at x miles per hr for entire trip.

40 = x T

T=40/x

So

$\dfrac{t+t'}{T}=\dfrac{32+0.2y}{40}$

To find numerical value of above expression we have to know the value of y.

8 0

4 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

Oksi-84 [34.3K]

Answer:

the equation of motion is

$x(t) =-\frac{1}{6} \cos4\sqrt{6} t$

Explanation:

Given that,

The weight attached to the spring is 24pounds

Acceleration due to gravity is 32ft/s²

Assume x is the string length, 4inches

convert the length inches to to feet = 4/12 = 1/3feet

From Hookes law , we calculate the spring constant k

k = W / x

k = 24 / (1/3)

k = 24 / 0.33

k = 72lb/ft

If the mass is displace from its equilibrium position by amount x

the differential equation is

$m\frac{d^2x}{dt^2} + kx=0\\\\\frac{3}{4 } \frac{d^2x}{dt^2}+72x=0\\$

$\frac{d^2x}{dt^2} +96x=0$

Auxiliary equation is

$m^2+96=0\\m=\sqrt{-96} \\m=\_ ^+4\sqrt{6}$

Thus the solution is,

$x(t) = c_1cos4\sqrt{6t} +c_2sin\sqrt{6t}$

$x'(t) =-4\sqrt{6c_1} sin4\sqrt{6t} +c_24\sqrt{6} cos4\sqrt{6t}$

The mass is release from rest

x'(0) = 0

$-4\sqrt{6c_1 } \sin4\sqrt{6} (0)+c_24\sqrt{6} \cos4\sqrt{6} (0)=0\\c_24\sqrt{6} =0\\\\c_2=0$

Therefore

x(t) = c₁ cos4 √6t

x(0) = -2inches

c₁ cos4 √6(0) = 2/12feet

c₁= 1/6feet

There fore, the equation of motion is

$x(t) =-\frac{1}{6} \cos4\sqrt{6} t$

7 0

4 years ago