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lukranit [14]
3 years ago
9

Train A is moving at 100 kmh–1 through a station. Train B is also travelling at 100 kmh–1 through the station but in the opposit

e direction to train A. What is the speed of a seated passenger on? (a) train A relative to an observer on the station? (b) train A relative to another seated passenger on the same train? (c) train B relative to a passenger on train A?
Physics
1 answer:
lisov135 [29]3 years ago
7 0

Explanation:

(a) An observer on the station has a speed of 0 km/h.  The speed of  a passenger on Train A is 100 km/h.  The relative speed is 100 km/h − 0 km/h = 100 km/h.

(b) The speed of both passengers is 100 km/h, in the same direction.  The relative speed is 100 km/h − 100 km/h = 0 km/h.

(c) The speed of both passengers is 100 km/h, in opposite directions.  The relative speed is 100 km/h − (-100 km/h) = 200 km/h.

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3 years ago
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What is an independent variable
snow_tiger [21]

Answer:

A variable (often denoted by x ) whose variation does not depend on that of another.

Explanation:

8 0
3 years ago
The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by t
stich3 [128]

Answer:

\dfrac{dV}{dt} = -0.466 m^3/s

Explanation:

given,

P (in kilo pascals),         volume V (in liters),        temperature T (in kelvins)

P V = 8.31 T

Rate of increase the temperature =  0.1 K/s

temperature = 285 K

Pressure = 18 kPa

increasing at the rate of 0.07 k Pa/s

Rate at which volume is changing = ?

V = 8.31 \dfrac{T}{P}

\dfrac{dV}{dt} = 8.31 \dfrac{P\dfrac{dT}{dt}-T\dfrac{dP}{dt}}{P^2}

\dfrac{dV}{dt} = 8.31 \dfrac{18 \times 0.1-285\times 0.07}{18^2}

\dfrac{dV}{dt} = 8.31 \dfrac{-18.15}{324}

\dfrac{dV}{dt} = -0.466 m^3/s

7 0
3 years ago
The launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m, where
lora16 [44]

Answer:

1.6 m

Explanation:

Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.

The time for landing should be calculated by using the second equation of motion formula

h = Ut + 1/2gt^2

Let U = 0

0.5 = 1/2 × 9.8 × t^2

0.5 = 4.9t^2

t^2 = 0.5 / 4.9

t^2 = 0.102

t = 0.32 s

The target should be placed so that the toy car lands on it at:

Distance = 5 × 0.32

distance = 1.597 m

Distance = 1.6 m

Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.

7 0
3 years ago
Jan first uses a Michelson interferometer with the 606 nmnm light from a krypton-86 lamp. He displaces the movable mirror away f
gogolik [260]

Answer:

a) d₁ = 247.8 μm

d₂ = 205.3 μm

b) d₂ = 20.53 x 10⁻⁵ m = 205.3 μm

Explanation:

a)

The formula for Michelson Interferometer is derived to be:

d = mλ/2

where,

d = distance moved

m = no. of fringes

λ = wavelength of light

For JAN, we have following data

d = d₁

m = 818

λ = 606 nm = 606 x 10⁻⁹ m

Therefore,

d₁ = (818)(606 x 10⁻⁹ m)/2

<u>d₁ = 24.78 x 10⁻⁵ m = 247.8 μm</u>

For LINDA, we have following data

d = d₂

m = 818

λ = 502 nm = 502 x 10⁻⁹ m

Therefore,

d₂ = (818)(502 x 10⁻⁹ m)/2

<u>d₂ = 20.53 x 10⁻⁵ m = 205.3 μm</u>

b)

The resultant displacement can be found out from the difference between both displacement. And the direction of resultant displacement will be the same as the direction of greater displacement. Therefore,

Resultant Displacement = Δd = d₁ - d₂

Δd = 247.8 μm - 205.3 μm

<u>Δd = 42.5 μm (in the direction of JAN)</u>

4 0
3 years ago
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