Two stars orbit each other in a

Relativistic kinetic energy is conserved in all reference frames. True O False

Free_Kalibri [48]

Answer:

i think it is true

Explanation:

7 0

3 years ago

An electron is projected with an initial speed vi = 4.60 × 105 m/s directly toward a very distant proton that is at rest. Becaus

frutty [35]

Answer:

$2.99\times 10^{-19}\ m$

Explanation:

<u>Given:</u>

$u$ = initial velocity of the electron = $4.60\times 10^5\ m/s$

$v$ = final velocity of the electron = $3u$

$x$ = initial position of the electron from the proton = very distant = $\infty$

<u>Assume:</u>

$m$ = mass of an electron = $9.1\times10^{-31}\ kg$

$e$ = magnitude of charge on an electron = $1.6\times10^{-19}\ C$

$p$ = magnitude of charge on an proton = $1.6\times10^{-19}\ C$

$k$ = Boltzmann constant = $9\times 10^9\ Nm^2/C^2$

$y$ = final position of the electron from the proton

$\Delta K$ = change in kinetic energy of the electron

$W$ = work done by the electrostatic force

$F$ = electrostatic force

$r$ = instantaneous distance of the electron from the proton

Let us first calculate the work done by the electrostatic force.

$W=\int Fdr\\\Rightarrow W = \int \dfrac{kep}{r^2}dr\\\Rightarrow W = kep\int \dfrac{1}{r^2}dr\\\Rightarrow W = kep\left | \dfrac{1}{r} \right |_{y}^{x}\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{y} \right )\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{\infty} \right )\\\Rightarrow W =\dfrac{kep}{x}$

Using the principle of the work-energy theorem,

As only the electrostatic force is assumed to act between the two charges, the kinetic energy change of the electron will be equal to the work done by the electrostatic force on the electron due to proton.

$\therefore \Delta K = W\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m((3u)^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m(8u^2)= \dfrac{kep}{x}\\\Rightarrow x= \dfrac{2kep}{8mu^2}\\\Rightarrow x= \dfrac{2\times 9\times 10^9\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{8\times 9.1\times10^{-31}\times (4.60\times 10^5)^2}\\\Rightarrow x=2.99\times 10^{-10}\ m\leq$

Hence, the electron is at a distance of $2.99\times 10^{-10}\ m$ when the electron instantaneously has speed of three times the initial speed.

8 0

4 years ago

A group of lions is chasing a zebra. The lions are most likely responding to which kind of stimulus

uranmaximum [27]

Answer:

The external stimulus of fear.

Explanation:

The zebras would most likely be scared of the lions, due to the fact that the lions want to eat the zebra. Thus leading to the zebra's running away from the lions.

5 0

3 years ago

A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the lad

pantera1 [17]

Answer: 45.3°

Explanation:

Given,

Length of ladder = l

Weight of ladder = w

Coefficient of friction = μs = 0.495

Smallest angle the ladder makes = θ

If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,

N1 = mg and

N2 = μmg

Moment at a point A in the clockwise direction is

N2 Lsinθ - mg.(L/2).cosθ = 0

μmgLsinθ - mg.(L/2).cosθ = 0

μmgLsinθ = mg.(L/2).cosθ

μsinθ = cosθ/2

sin θ / cos θ = 1 / 2μ

Tan θ = 1 / 2μ

Substituting the value of μ = 0.495, we have

Tan θ = 1 / 2 * 0.495

Tan θ = 1 / 0.99

Tan θ = 1.01

θ = tan^-1(1.01)

θ = 45.3°

3 0

3 years ago

A pitcher throws an overhand fastball from an approximate height of 2.65 m and at an angle of 2.5° below horizontal. The catcher

rodikova [14]

Answer:

The initial velocity of the pitch is approximately 36.5 m/s

Explanation:

The given parameters of the thrown fastball are;

The height at which the pitcher throws the fastball, h₁ = 2.65 m

The angle direction in which the ball is thrown, θ = 2.5° below the horizontal

The height above the ground the catcher catches the ball, h₂ = 1.02 m

The distance between the pitcher's mound and the home plate = 18.5 m

Let 'u' represent the initial velocity of the pitch

From h = $u_y$ ·t + 1/2·g·t², we have;

$u_y$ = The vertical velocity = u·sin(θ) = u·sin(2.5°)

h = 2.65 m - 1.02 m = 1.63 m

uₓ·t = u·cos(θ) = u·cos(2.5°) × t = 18.5 m

∴ t = 18.5 m/(u·cos(2.5°))

∴ h = $u_y$ ·t + 1/2·g·t² = (u·sin(2.5°))×(18.5/(u·cos(2.5°))) + 1/2·g·t²

1.63 = 8.5·tan(2.5°) + 1/2 × 9.8 × t²

t² = (1.63 - 8.5·tan(2.5°))/(1/2 × 9.8) = 0.25691469087

t = √(0.25691469087) ≈ 0.50686752763

t ≈ 0.50686752763 seconds

u = 18.5 m/(t·cos(2.5°)) = 18.5 m/(0.50686752763 s × cos(2.5°)) = 36.5334603 m/s ≈ 36.5 m/s

The initial velocity of the pitch = u ≈ 36.5 m/s.

3 0

3 years ago

Two stars orbit each other in a _ solar system