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Arada [10]
3 years ago
7

A futuristic train rushes past you at an incredible, relativistic speed such that it appears to only be 80% as long as you know

it to be. At what speed is that train traveling from the point of view of your reference frame
Physics
1 answer:
Fed [463]3 years ago
3 0

Answer:

v = 0.6c = 1.8 x 10⁸ m/s

Explanation:

From Einstein's theory of relativity, we know that the length of an object contracts while traveling at a speed relative to speed of light. The contraction is according to following equation:

L = L_{0}\sqrt{1 - \frac{v^2}{c^2}}

where,

L = Relative Length

L₀ = Rest Length

According to given:

L = 0.8 L₀

c = speed of light = 3 x 10⁸ m/s

v = relative speed of train = ?

Therefore,

0.8L_{0} = L_{0}\sqrt{1 - \frac{v^2}{c^2} } \\\\0.8 = \sqrt{1 - \frac{v^2}{c^2} }

squaring on both sides:

0.64 = 1-\frac{v^2}{c^2}\\\\\frac{v^2}{c^2} = 1-0.64 \\v^2= 0.36c^2\\

taking square root on both sides:

<u>v = 0.6c = 1.8 x 10⁸ m/s</u>

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A cup containing 200 g of hot water is taken off the stove placed on the kitchen table. Initially the water is at 75Degree C. bu
8090 [49]

Answer:

ΔH = -45.1872 kJ , where negative sign signifies heat loss.

Q = ΔH = -45.1872 kJ  , where negative sign signifies heat loss.

S system = -0.141 kJ/K

S surroundings = 0.1536 kJ/K

S universe = -0.141 kJ/K + 0.1536 kJ/K  = 0.0126 kJ/K

Explanation:

Given:

Cp = 4. 184 J/(mole. K)

T₁ = 75 ⁰C

T₂ = 21 ⁰C

Mass of water = 200 g = 0.2 kg

Since,

\Delta H=m\times C\times (T_f-T_i)

ΔH = 0.2*4.184*(21-75)  kJ

<u> ΔH = -45.1872 kJ , where negative sign signifies heat loss.</u>

Since the process is at constant pressure

<u> Q = ΔH = -45.1872 kJ  , where negative sign signifies heat loss.</u>

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

T₁ = 75 ⁰C = 348 .15 K

T₂ = 21 ⁰C = 294.15 K

The entropy of the water is given by:

<u> S = m×Cp×ln(T₂ /T₁) </u>

S = 0.2*4.184*ln(294.15/348.15)

<u> S system = -0.141 kJ/K</u>

The heat gain by surroundings

<u> dQ = -Qreaction =  45.1872 kJ </u>

The entropy change of surroundings is

S surr = dQ/T₂ = 45.1872/294 .15

<u> S surr = 0.1536 kJ/K </u>

The entropy of universe  is the sum total of the entropy of the system and the surroundings and thus,

S universe = Ssys + Ssurr

<u> S universe = -0.141 kJ/K + 0.1536 kJ/K  = 0.0126 kJ/K</u>

4 0
3 years ago
A cylindrical bucket, open at the top, is 28.0 cm high and 11.0 cm in diameter. A circular hole with a cross-sectional area 1.55
svlad2 [7]

Answer:

so height is 0.1283 m

Explanation:

given data

height = 28 cm

diameter = 11 cm

cross-sectional area = 1.55 cm2

water flow rate  =  2.46×10^−4 m3/s

to find out

How high will the water in the bucket rise

solution

we know that here

potential energy = kinetic energy

mgh = 1/2 mv²

multiply both sides by the 2 and we get

2mgh=mv²

solve it we get

√(2gh) = v    ....................1

h = v²/2g   ...............2

and

flow rate = A V

2.46×10^−4 = V 1.55×10^−4

V = 1.5870 m/s

so from 2

h = v²/2g

h = 1.5870²/ 2(9.81)

h = 0.1283 m

so height is 0.1283 m

6 0
3 years ago
you are given a 1/5 model of a car that normally runs at 30 mph. what wind tunnel speed should you use to test the car? (assume
julsineya [31]

The wind tunnel speed that must be used to test the car 1 / 5 model of a car that normally runs at 30 mph is 150 mph

Re = u L / ν

Re = Reynold's number

u = Flow speed

L = Characteristic linear dimension

ν = Kinematic viscosity

ν_{p} = 1.516 * 10^{-5} m² / s

ν_{p} = ν_{m} = 1.516 * 10^{-5} m² / s

L_{m} = L_{p} / 5

u_{p} = 30 mph

Re_{m} = u_{m} L_{p} / 5 * 1.516 * 10^{-5}

Re_{p} = 30 L_{p} / 1.516 * 10^{-5}

In tunnel,

Re_{m} = Re_{p}

u_{m} L_{p} / 5 * 1.516 * 10^{-5} = 30 L_{p} / 5 * 1.516 * 10^{-5}

u_{m} = 30 * 5

u_{m} = 150 mph

Therefore, the wind tunnel speed that must be used to test the car is 150 mph

To know more about Reynold's number

brainly.com/question/12977616

#SPJ1

5 0
1 year ago
The sun is 60° above the horizon. Rays from the sun strike the still surface of a pond and cast a shadow of a stick that is stuc
Kipish [7]

Answer:

shadow length 7.67 cm

Explanation:

given data:

refractive index of water is 1.33

by snell's law we have

n_{air} sin30 =n_{water} sin\theta

1*0.5 = 1.33*sin\theta

solving for\theta

sin\theta = \frac{3}{8}

\theta = sin^{-1}\frac{3}{8}

\theta =  22 degree

from shadow- stick traingle

tan(90-\theta) = cot\theta = \frac{h}{s}

s = \frac{h}{cot\theta} = h tan\theta

s = 19tan22 = 7.67 cm

s = shadow length

5 0
3 years ago
If given both the speed of light in a material and an incident angle. How can you find the refracted angle?
Basile [38]

Answer:

r = Sin^{-1}\left ( \frac{v Sini}{c} \right )

Explanation:

Let the speed of light in vacuum is c and the speed of light in medium is v. Let the angle of incidence is i.

By using the definition of refractive index

refractive index of the medium is given by

n = speed of light in vacuum / speed of light in medium

n = c / v  ..... (1)

Use Snell's law

n = Sin i / Sin r

Where, r be the angle of refraction

From equation (1)

c / v = Sin i / Sin r

Sin r = v Sin i / c

r = Sin^{-1}\left ( \frac{v Sini}{c} \right )

8 0
4 years ago
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