A closed system is BEST described as

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea

german

Answer:

Q1) Time taking by particle to travel the 40 km wrt. earth = $1.34\times10^{-6}$ sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = $1.285\times10^{-5}$ sec.

Explanation:

Given :

Speed of particle wrt. earth $v=0.99537c$

Distance between where particle is created and earth surface = 40 km

we know that,

⇒ $v = \frac{x}{t}$

Where $x = 40\times10^{3} m$ , $v = 0.99537c$ , we know speed of light $c = 3 \times10^{8}$

∴ $t = \frac{x}{v}$

$= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }$

$t = 1.34\times10^{-6} sec$

∴ Thus, time taking by particle to travel the 40 km wrt. earth $t = 1.34\times10^{-6} sec$

According to the lorentz transformation,

⇒ $l = l_{o} \sqrt{1-\frac{v^2}{c^2} }$

Where $l =$ improper length, $l_{o} =$ proper length (distance measured wrt. rest frame) = 40 km

$l = 40 \sqrt{1-\frac{v^2}{c^2 }$

$l = 40 \times 0.096$

$l = 3.84$ km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

$\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }$

Where $\Delta t =$ improper time (wrt. earth frame time) $=1.34\times10^{-6} sec$ , $\Delta t _{o} =$ proper time (wrt. particle frame).

$1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}$

$\Delta t_{o} = 1.285 \times10^{-5} sec$

Thus, the time taking by particle to travel from where it is created to the surface of the earth $= 1.285 \times10^{-5} sec$ .

6 0

2 years ago

A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius

disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

KE = 0.23 J

given parameters

Disk radius R = 15 cm = 0.15 m
Cylinder radius r = 1.5 cm = 0.0015 m
Disk mass M = 20 kg
Time t = 1.2 s
Force F = 12 N

to find

Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk I_{disk} = ½ M R²

cylinder I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

I_{total} = ½ (M R² + m r²) = ½ M R² ( $1 + \frac{m}{M} \ (\frac{r}{R})^2$ )

I_{total} = ½ M R² ( $1+ \frac{m}{20} (\frac{0.015}{0.15} )^2$ ) = $\frac{1}{2}$ M R² (1 + 0.005 m)

As the shaft mass is much lighter than the disk mass , the last term is very small, which is why we despise it.

I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

τ = I α

α = $\frac{\tau }{I_{total}}$ l

τ = F R

α = $\frac{F \ R}{I_{total}}$

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

w = w₀ + α t

where w is the angular velocity, alpha is the angular acceleration and t is the time

w = 0 + $\frac{\tau }{I_{total}} \ t$

we substitute in the kinetic energy equation

KE = ½ I_{total} ( $\frac{ \tau }{I_{total}} \ t$ )²

KE = ½ $\frac{ \tau^2 }{I_{total}} \ t^2$

let's substitute

KE = $\frac{F^2 \ R^4}{M \ R^2 } \ t^2$

KE = F² R² t² / M

let's calculate

KE = 12² 0.15² 1.2² / 20

KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

KE = 0.23 j

learn more about rotational kinetic energy here:

brainly.com/question/20261989

4 0

2 years ago

If 28.0 J of work is done by an external force to move a charge from a potential of 8.0 V 8.0 V to a potential of 4.0 V , 4.0 V,

jonny [76]

Answer:

Change in electric potential energy is -28.0 J

Explanation:

Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.

OR

Electric potential energy is also equal to the change in the configuration of the charge particles.

Thus,

Change in electric potential energy = - Work Done

According to the problem, Work Done is equal to 28 J. Thus,

Change in electric potential energy = -28 J

5 0

3 years ago

a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T

wlad13 [49]

Explanation:

Given: $F=m\ddot{x}=Fe^{-\frac{t}{T}}$

Solving for $\ddot{x}$ :

$\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}$

where:

$T=\sqrt{\frac{m}{F}}$

Integrating to get $\dot{x}$ with initial conditions $\dot{x}(0)=0$ :

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}$

Integrating to get x with initial conditions x(0) = 0:

$x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}$

When t=T:

$x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}$

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})$

4 0

2 years ago

Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni

Ber [7]

Answer:

a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate, c) x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

F = m a

the electric force is

F = q E

we substitute

q E = m a

a = qE / m

the mass of the particle is m = 2.00 10-16 kg

a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

a = 1,616 m / s²

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

sin θ = v_{oy} / v

cos θ = v₀ₓ / v

v_{oy} = v₀ sin θ

v₀ₓ = v₀ cos θ

v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

0 = 0 + v_{oy} t - ½ a_y t²

v_{oy}= ½ a_y t

t = 2v_{oy} / a_y

let's calculate

t = 2 0.6139 10⁵ / 1.616

t = 7.597 10⁴s

in this time the particle travels a horizontal distance

x = v₀ₓ t

x = 0.8145 10⁵ 7.597 10⁴

x = 6.19 10⁹ m

the particle falls off the plate

4 0

2 years ago