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3 years ago

Compare the strong and weak megnetic field

Physics

1 answer:

garri49 [273]3 years ago

7 0

The strong magnetic fields is Long rang attractive power Reuther then week magnetic field.

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Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when t

yawa3891 [41]

Answer:

a) $x=2*10^{9} m$ and $t=8.35 s$

b) t = γt', so it is 8.35 s.

Explanation:

a) The equation of Lorentz transformations is given by:

$x=\gamma(x'+ut')$

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

x' = 0
t' = 5.00 s
u =0.800 c, c is the speed of light 3*10⁸ m/s

$\gamma=\frac{1}{\sqrt{1-(u/c)^{2}}}$

$\gamma=\frac{1}{\sqrt{1-(0.800c/c)^{2}}}$

$\gamma=\frac{1}{\sqrt{1-(0.800)^{2}}}$

$\gamma=1.67$

$x=1.67(0+0.800c*5.00)$

$x=2*10^{9} m$

Now, to find t we apply the same analysis:

$t=\gamma(t'+\frac{ux'}{c^{2}})$

but as x'=0 we just have:

$t=\gamma(t')$

$t=1.67*5.00=8.35 s$

b) Here, Mavis reads 5 s on her watch and Stanley measured the events at a time affected by the Lorentz factor, in other words t = γt', if we see it is the same a) part. So the time interval will be equal to 8.35 s.

I hope it helps you!

5 0

3 years ago

Read 2 more answers

A student walks 1.0 mi west and then 1.0 mi north. afterward, how far is she from her starting point?

svetlana [45]

1.0 mi because as he has covered 1 the disrance of 1 m and then 1 m in north so he can go straight 1m north so the distance is actually 1m from the starting point

7 0

3 years ago

A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Fin

Ira Lisetskai [31]

Answer:

a) d = (6.00 t i ^ + 0.500 t²) m , b) v = (6.00 i ^ + 1.00 t j ^) m / s

c) d = (24.00 i ^ + 8.00 j^ ) m , d) v = (6.00 i ^ + 5 j^ ) m/s

Explanation:

This exercise is about kinematics in two dimensions

a) find the position of the particle on each axis

X axis

Since there is no acceleration on this axis, we can use the relation of uniform motion

v = x / t

x = v t

we substitute

x = 6.00 t

Y Axis

on this axis there is an acceleration and there is no initial speed

y = v₀ t + ½ a t²

y = ½ at t²

we substitute

y = ½ 1.00 t²

y = 0.500 t²

in vector position is

d = x i ^ + y j ^

d = (6.00 t i ^ + 0.500 t²) m

b) x axis

as there is no relate speed is concatenating

vₓ = v₀

vₓ = 6.00 m / s

y Axis

there is an acceleration and the initial speed is zero

$v_{y}$ = v₀ + a t

v_{y} = a t

v_{y} = 1.00 t

the velocity vector is

v = vₓ i ^ + v_{y} j ^

v = (6.00 i ^ + 1.00 t j ^) m / s

c) the coordinates for t = 4 s

d = (6.00 4 i ^ + 0.50 4 2 j⁾

d = (24.00 i ^ + 8.00 j^ ) m

x = 24.0 m

y = 8.00 m

d) the velocity of for t = 4 s

v = (6 i ^ + 1 5 j ^)

v = (6.00 i ^ + 5 j^ ) m/s

7 0

3 years ago

Which term describes an experiment with one dependent variable and one experimental variable

wlad13 [49]

A. Controlled experiment hope this helps

8 0

3 years ago

Read 2 more answers

Which of the following is not a component of a normal experiment?

MaRussiya [10]

Answer:

D. Validity group

Explanation:

A validity group is not necessarily a component of a normal experiment.

A normal experiment is made up of the dependent variables, independent variables, and most times the control group.

The dependent variable is one that relies on the outcome of the effect produced from the experimental design
The independent variable is the one that causes a desired effect under the study.
The control group is a group that does not receive any treatment nor attenuation of the conditions.

4 0

3 years ago