Ask question

Ask question

All categories

3 years ago

6

A certain automobile is 6.0 m long if at rest. If it is measured to be 4.8 m long while moving, its speed is:

1 answer:

tatuchka [14]3 years ago

7 0

Answer:

1.8*10^8m/s

Explanation:

Using

L= lo√1-v²/c²

So making v subject we have

V= c√1-4.8²/6²

V= 0.6*c

V= 0.6*3E8m/s

V= 1.8*10^8m/s

You might be interested in

Um objeto sobe verticalmente com velocidade de 10m/s . Quando sua altura é de 60m , uma parte desse objeto cai . Considerando g=

schepotkina [342]

Why is it in Spanish????????????????

6 0

3 years ago

How are humans able to perceive light waves? Sound waves?

Vesna [10]

Sound and light are similar in that both are forms of energy that travel in waves. They both have properties of wavelength, freqency and amplitude. ... Sound is a form of mechanical energy caused by vibrations of matter. Light is electromagnetic energy caused by interacting electric and magnetic fields.

7 0

3 years ago

Use the component form of newton's second law to write an expression for the x component of the net force, σfx

nordsb [41]

Draw a free body diagram to show which forces act in the x and y directions. The x component equation is σfx = 0. The σfx being all the forces acting in the x direction.

8 0

3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

If a car travels 30 mi. north for 30 min., 60 mi. east for 1.0 hour, and 30 mi. south for 30 min., what is the average speed

Ainat [17]

Answer: 60mph

Explanation:

Given the following :

First leg travel:

Distance = 30 miles

Time of travel= 30 minutes = 0.5 hour

Second leg travel:

Distance = 60 miles

Time of travel = one hour

Average speed :

Speed = total Distance / time of travel

Total distance in miles = (30 + 60) miles = 90 miles

Total time of travel = 1 hour + 0.5 hour = 1.5 hours

Average speed = total distance traveled / total travel time

Average speed = 90 miles / 1.5 hours

Average speed = 60 miles / hour

= 60mph

3 0

3 years ago