تقطع اولا مسافة 8 km شمالا من البيت ثم تمشي شرقا حتى تكون ازاحتك من البيت 10km ما مقدار المسافة التي قطعتها شرقا

As an object moves, the distance it travels increases with time. Agree Disagree

Mnenie [13.5K]

Agree

Hope this helps

4 0

2 years ago

Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measur

Over [174]

Answer:

the magnitude of the velocity of one particle relative to the other is 0.9988c

Explanation:

Given the data in the question;

Velocities of the two particles = 0.9520c

Using Lorentz transformation

Let relative velocity be W, so

v $_r$ = ( u + v ) / ( 1 + ( uv / c²) )

since each particle travels with the same speed,

u = v

so

v $_r$ = ( u + u ) / ( 1 + ( u×u / c²) )

v $_r$ = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )

we substitute

v $_r$ = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )

v $_r$ = 1.904c / ( 1 + 0.906304 )

v $_r$ = 1.904c / 1.906304

v $_r$ = 0.9988c

Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c

5 0

2 years ago

What distance does light travel in water, glass, and diamond during the time that it travels 1.0 m in vacuum? The refractive ind

lidiya [134]

Answer:

refractive index for water,glass,diamond are 0.752m, 0.667m, 0.413m respectively

Explanation:

refractive index (n) = $\frac{velocity of light in air/vacuum}{velocity of light in substance}$

velocity = $\frac{distance}{time}$

The time for travel is kept constant for all mediums.

refractive index (n) = $\frac{\frac{distance in vacuum}{time} }{\frac{distance in medium}{time} }\\ \\=\frac{distance in vacuum}{distance in medium}$

distance in medium = $\frac{distance in vacuum}{refractive index of medium}$

$S_{medium} = \frac{S_{vacuum} }{n_{medium} }$

For water, n= 1.33

$S_{water} = \frac{1 }{1.33}$

$S_{water} = 0.752m$

For glass, n=1.5

$S_{glass}= \frac{1 }{1.5}$

$S_{glass} = 0.667m$

For diamond, n= 2.42

$S_{diamond} = \frac{1 }{2.42}$

$S_{diamond} = 0.413m$

3 0

2 years ago

Pweese help one more timeeee pweese look at the image below

4vir4ik [10]

Answer:

increasing; speeding up is my answer

7 0

2 years ago

Read 2 more answers

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

RoseWind [281]

Answer:

a) $\Delta U_{g} = 12.945\,J$ , b) $\Delta U_{k} = 12.945\,J$ , c) $k = 2930.059\,\frac{N}{m}$

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

4 0

2 years ago