Ask question

Ask question

All categories

3 years ago

12

A tourist is walking at a speed of 1.3 m/s along a 9.0-km path that follows an old canal. If the speed of light in a vacuum were

3.0 m/s, how long would the path be, according to the tourist?

2 answers:

Marat540 [252]3 years ago

8 0

Answer:

The length is 8111.1 m.

Explanation:

Given that,

Speed of tourist = 1.3 m/s

Distance = 9.0 km

Speed of light = 3.0 m/s

We need to calculate the length

Using formula of length

$L= L_{0}\sqrt{1-(\dfrac{v}{c})^2}$

Where, l = length

$L_{0}$ =original length

v = speed of tourist

c = speed of light

Put the value into the formula

$L=9000\times\sqrt{1-(\dfrac{1.3}{3.0})^2}$

$L=8111.1\ m$

Hence, The length is 8111.1 m.

Alecsey [184]3 years ago

5 0

Answer:

The length of path according to the tourist is 8111.10 meters.

Explanation:

It is given that,

Speed of tourist, v = 1.3 m/s

Path length, L₀ = 9 km = 9000 m

If the speed of light in a vacuum were 3.0 m/s. We need to find the length of path according to the tourist. Let it is L. It is a case of length contraction.

So, $L=L_0\sqrt{1-\dfrac{v^2}{c^2}}$

$L=9000\times \sqrt{1-\dfrac{(1.3)^2}{(3)^2}}$

L = 8111.10 m

So, the length of path according to the tourist is 8111.10 meters. Hence, this is the required solution.

You might be interested in

Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face

Sunny_sXe [5.5K]

Answer:

$\mu = 1.645$

Explanation:

By Snell's law we know at the left surface

$\theta_i = 19^o$

$\theta_r = ?$

$\mu_1 = 1$

$\mu_2 = \mu$

now we have

$1 sin19 = \mu sin\theta_r$

$0.33 = \mu sin\theta_r$

now on the other surface we know that

angle of incidence = $\theta_r'$

$\theta_e = 90$

so again we have

$\mu sin\theta_r' = 1 sin90$

so we have

$\theta_r = sin^{-1}\frac{0.33}{\mu}$

$\theta_r' = sin^{-1}\frac{1}{\mu}$

also we know that

$\theta_r + \theta_r' = 49$

$sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49$

By solving above equation we have

$\mu = 1.645$

3 0

3 years ago

Read 2 more answers

A student was trying to find the relationship between mass and force. He placed four different masses on a table and pulled them

Gwar [14]

Answer:

B. There is a direct proportion between the mass and force listed in the table.

Explanation:

From the table, the values of force increases with increase in the value of mass.

if 5kg=25 N

Finding the contant of proportionality k;

k=25/5=5

thus M=k(F)...........where M is mass in kg and F is force in newton, then

M=5F

This show that for every value of mass, we get the value of Force if we multiply by a contant k=5

This means there is a direct proportionality relation between mass and force in the table.

5 0

3 years ago

Read 2 more answers

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago

Define 1 pascal pressure

Rasek [7]

Answer:For example, standard atmospheric pressure (or 1 atm) is defined as 101.325 kPa. The millibar, a unit of air pressure often used in meteorology, is equal to 100 Pa. (For comparison, one pound per square inch equals 6.895 kPa.)

Explanation:A pascal is a pressure of one newton per square metre, or, in SI base units, one kilogram per metre per second squared.

I hope this helps.... I'm sorry if it doesn't

5 0

2 years ago

Please anyone post any class 9 physics question

aalyn [17]

Answer:

I will but can you just wait for some minutes cus I am in a hurry now.

sorry that pic is a little blurry

6 0

2 years ago