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Oksi-84 [34.3K]

4 years ago

8

Suppose you are in a spaceship traveling at 99% of the speed of light past a long, narrow space station. Your direction of trave

l is parallel to the length of the station. If you measure lengths of objects on the station and also how time is passing on the station, what results will you get?
A) Lengths will appear shorter and time will appear to pass slower.
B) Lengths will appear shorter and time will appear to pass faster.
C) Lengths will appear shorter and time will appear to pass faster.

2 answers:

satela [25.4K]4 years ago

5 0

Answer:

A. Lengths will appear shorter and time will appear to pass slower.

Explanation:

From theory of relativity, we know that:

L₀ = length of object, measured in stationary frame of reference

L = length of object measured from a frame moving with respect to object, called ‘relativistic length’

v = relativistic speed between observer and the object

c = speed of light

then,

L = L₀ √(1-v²/c² )

Hence, the length of the object decreases with the increase in its relativistic speed "v"

t₀ = time measured by clock at rest with respect to event.

t = time measured by clock in motion relative to the event

v = relativistic speed between observer and the object

c = speed of light

then,

t = t₀/√(1-v²/c² )

Hence, the time increases with the increase in in relativistic speed and as a result it appears to pass slower.

Hence, the correct option is:

<u>A. Lengths will appear shorter and time will appear to pass slower.</u>

iVinArrow [24]4 years ago

3 0

Answer:

B) Lengths will appear shorter and time will appear to pass faster.

Explanation:

This is in line with the laws of relativity.

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A toroidal solenoid has 600 turns, cross-sectional area 6.90 cm2, and mean radius 4.30 cm.

Dahasolnce [82]

(a) The coil's self-inductance is 7.26 mH.

(b) The self-induced emf in the coil is 7.26 V

(c) The direction of the induced emf is from b to a.

<h3>Coil's self-inductance</h3>

L = N²μA/I

L = (600² x 4π x 10⁻⁷ x 6.9 x 10⁻⁴)/(0.043)

L = 7.26 x 10⁻³ H

L = 7.26 mH

<h3>Self-induced emf in the coil</h3>

emf = N(ΔBA)/t

where;

B is magnetic field
A is area
N is number of turns
t is time

B = μNI/L

B1 = (4π x 10⁻⁷ x 600 x 5)/0.043

B1 = 0.0876 T

B2 = (4π x 10⁻⁷ x 600 x 2)/0.043

B2 = 0.035 T

emf = NΔBA/t

emf = (600)(0.0876 - 0.035)(6.9 x 10⁻⁴) / (3 x 10⁻³)

emf = 7.26 V

The direction of the induced emf is always opposite to the direction of the applied current.

Thus, the direction of the induced emf is from b to a.

Learn more about induced emf here: brainly.com/question/13744192

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2 years ago

Why does mars take 462.3 earth days more than venus to relove around the sun?

steposvetlana [31]

Answer:

c.The sun is closer to vinus than it is to mars

Explanation:

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2 years ago

Read 2 more answers

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses

Tom [10]

Answer:

a) U = 0.375 mJ

b) p_rad = 4.08 mPa

c) λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

d) E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

Explanation:

Given:

λ_air : wavelength = 810 * 10^(-9) m

P: Power delivered = 0.25 W

d : Diameter of circular spot = 0.00051 m

c : speed of light vacuum = 3 * 10^8 m/s

n_air : refraction Index of light in air = 1

n_med : refraction Index of light in medium = 1.34

ε_o : permittivity of free space = 8.85 * 10^-12 C / Vm

part a

The Energy delivered to retina per pulse given that laser pulses are 1.50 ms long:

U = P*t

U = (0.25 ) * (0.0015 )

U = 0.375 mJ

Answer : U = 0.375 mJ

part b

What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed?

p_rad = I / c

Where I : Intensity = P / A

p_rad = P / A*c

Where A : Area of circular spot = pi*d^2 / 4

p_rad = 4P / pi*d^2*c

p_rad = 4(0.25) / pi*0.00051^2*(3.0 * 10^8)

p_rad = 0.00408 Pa

Answer : p_rad = 4.08 mPa

part c

What are the wavelength and frequency of the laser light inside the vitreous humor of the eye?

λ_med = n_air*λ_air / n_med

λ_med = (1) * (810 nm) / 1.34

λ_med = 604 nm

f_med = f_air

f_med = c / λ_air

f_med = (3*10^8) / (810 * 10^-9)

f_med = 3.7 * 10^14 Hz

Answer : λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

d)

What is the electric and magnetic field amplitude in the laser beam?

I = P / A

I = 0.5*ε_o*c*E_o ^2

I = 4P / pi*d^2

Hence, E_o = ( 8 P / ε_o*c*pi*d^2 ) ^ 0.5

E_o = ( 8 * 0.25 / (8.85*10^-12) * (3*10^8) * π * (0.00051)^2) ^ 0.5

E_o = 3.04 * 10^4 V / m

For maximum magnetic field strength:

B_o = E_o / c

B_o = 3.04 * 10^4 / (3*10^8)

B _o = 1.013 *10^-4 Nm/Amp

Answer: E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

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3 years ago

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

siniylev [52]

Answer:

r = 1.86 m

Explanation:

Here the force due to wall of the cylinder is towards the axis of the cylinder

This force will act as centripetal force for the people sit inside the chamber

now we will have

$F_c = \frac{mv^2}{R}$

now we will have

$F_c = 488 N$

m = 84.4 kg

v = 3.28 m/s

now we have

$488 = \frac{84.4(3.28)^2}{r}$

now we have

$r = 1.86 m$

8 0

3 years ago

What is the force required to produce a 1.4 Nm torque when applied to a door at a 60.0º angle and 0.40 m from the hinge?

Marysya12 [62]

Refer to the diagram shown below.

The component of the applied force perpendicular to the door is
F * sin(60°) = 0.866F N

Because the moment arm is 0.40 m, the torque is
(0.866F N)*(0.4 m) = 0.3464F N-m

This torque is equal to 1.4 N-m, therefore
0.3464F = 1.4
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Answer: 4.04 N

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3 years ago