Answer:
182 to 3 s.f
Explanation:
Workdone for an adiabatic process is given as
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
where γ = ratio of specific heats. For carbon dioxide, γ = 1.28
For an adiabatic process
P₁V₁ʸ = P₂V₂ʸ = K
K = P₁V₁ʸ
We need to calculate the P₁ using ideal gas equation
P₁V₁ = mRT₁
P₁ = (mRT₁/V₁)
m = 2.80 g = 0.0028 kg
R = 188.92 J/kg.K
T₁ = 27°C = 300 K
V₁ = 500 cm³ = 0.0005 m³
P₁ = (0.0028)(188.92)(300)/0.0005
P₁ = 317385.6 Pa
K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
V₁ = 0.0005 m³
V₂ = 2.10 dm³ = 0.002 m³
1 - γ = 1 - 1.28 = - 0.28
W =
18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)
W = -67.47 (5.698 - 8.4)
W = 182.3 = 182 to 3 s.f
Well, first of all, the car is not moving at a uniform velocity, because,
on a curved path, its direction is constantly changing. Its speed may
be constant, but its velocity isn't.
The centripetal force on a mass 'm' that keeps it on a circle with radius 'r' is
F = (mass) · (speed)² / (radius).
For this particular car, the force is
(2,000 kg) · (25 m/s)² / (80 m)
= (2,000 kg) · (625 m²/s²) / (80 m)
= (2,000 · 625 / 80) (kg · m / s²)
= 15,625 newtons .
The gravitational force acting upon the woman is equal to <u>-588.6N</u>
Why?
To solve the problem, we need to consider that two forces are acting upon the woman, the first one is related to her weight and the second one is related to the acceleration of the elevator.
Gravitational force acting upon the woman:
Hence, we have that the gravitational force acting upon the woman is equal to -588.6N.
Have a nice day!
energetic electrically charged particles (mostly electrons) accelerate along the magnetic field lines into the upper atmosphere, where they collide with gas atoms, causing the atoms to give off light.
(exploratorium.com)
gotta cite those sources
Answer:
Explanation:
given,
scattering angle of alpha particle = 25.0° above its initial direction of motion
oxygen nucleus recoils at = 50.0° below this initial direction.
final speed of the oxygen = 2.08×10⁵ m/s
mass of alpha particle = 4.0 u
mass o oxygen nucleus = 16 u
momentum conservation along x- axis
....(1)
Along y-direction
putting value in equation (1)
