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andrew-mc [135]
3 years ago
15

A plane traveled south for 2.5 hrs at a velocity of 1200km/hr. What distance did it travel?

Physics
1 answer:
Oxana [17]3 years ago
6 0
From the formula, V=displacement/time
displacement/distance =velocity*time
distance=1200*2.5
distance=3*10³km
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3 years ago
The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it
Sindrei [870]

Answer with Explanation:

We are given that

Speed ,v_0=107257 kmph

Angular velocity,\omega=7.292\times 10^{-5} rad/s

Radius of earth,r=6371 km=6371000 m

1 km=1000 m

Linear velocity,v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s

Linear velocity,v=464.57\times \frac{18}{5}=1672.46 km/h

Velocity at point A,v_A=-vi+v_0 j=-1672.46 i+107257j kmph

Velocity at point B,v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph

Velocity at point C,v_C=v_0j+vi=1672.46 i+107257j kmph

Velocity at point  D,v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph

3 0
3 years ago
Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What p
Temka [501]

Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT...............................  Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

6 0
3 years ago
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