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3 years ago

14

2560 meters in 60 seconds, what is his average speed (velocity)

1 answer:

jok3333 [9.3K]3 years ago

5 0

42.6 is the answer I believe because you would do 2,560 divided by 60 if I'm correct.

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3 years ago

In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m

Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

$F = \dfrac{kq_1q_2}{r^2}$

$F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}$

F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a = 8.20 x 10⁻⁸

$a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}$

a = 9 x 10²² m/s²

c) speed of the electron

$a =\dfrac{v^2}{r}$

$9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}$

v² = 4.77 x 10¹²

v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}$

T = 1.53 x 10⁻¹⁶ s

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3 years ago

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Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr

DiKsa [7]

Answer:

(2/3) times height collision occur

Explanation:

for ball A

from the kinematic equation

the distance of ball A is

$x_A = v_0 t + \frac{1}{2} at^2$

$v_0* t$ = velocity ( time ) = distance

since ball is at height, the above equation changes as

$x_A = H - \frac{1}{2} gt^2$

for ball B

$xB = v0 t - \frac{1}{2} gt^2$

the condition of collision is

xA = xB

$vA = - 2vB$ (given)

from the kinematic equation

the speed of the ball A is

$v_A = u- gt$

since initial speed of the ball A is zero

, so

$v_A = -gt$

the speed of the ball B is

$v_B = v_0 - gt$

since $v_A = - 2v_B$

$-gt = -2 ( v_0 - gt)$

$-gt = -2 v_0 +2gt$

$3gt =2 v_0$

$t = \frac{2v_0}{3g}$

since $x_A = x_B$

$H - \frac{1}{2} gt^2= v_0 t - \frac{1}{2} gt^2$

$H = v_0 t$

$= v_0 (2v_0/3g)$

$= \frac{2 v_0^2}{ 3g}$

$x_A = 2 (\frac{v_0^2}{ 3g})- \frac{1}{2} gt^2$

$= 4 \frac{v_0^2}{9g} = (2/3) H$

so, (2/3) times height collision occur

3 0

3 years ago