When soccer players run they are using friction to propell themselves
An initial velocity is:
v o = 25 m/s
The vertical component of the initial velocity:
v o y = v o * sin 60° =
= v o * √3 / 2 = 25 m/s * √3 / 2 = 21.65 m/s
Answer:
The approximate vertical component of the initial velocity is 21.65 m/s.
Cloud Formation<span> Due to Surface Heating. Some </span>clouds<span>form due to the heating of the Earth's surface. First, the </span>Sun<span>heats the ground, which then heats the air. ... This extra water vapor begins to condense out of the air parcel in the form of liquid water droplets.</span>
Answer:
the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)
Explanation:
assuming ideal gas behaviour of the gas , the equation for ideal gas is
P*V=n*R*T
where
P = absolute pressure
V= volume
T= absolute temperature
n= number of moles of gas
R= ideal gas constant = 8.314 J/mol K
P=n*R*T/V
the work that is done by the gas is calculated through
W=∫pdV= ∫ (n*R*T/V) dV
for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:
W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)
since
P₁=n*R*T/V₁
P₂=n*R*T/V₂
dividing both equations
V₂/V₁ = P₁/P₂
W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )
replacing values
P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa
since P₂ = 1 atm = 101325 Pa
W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J
We have to calculate the impulse of a hockey puck.
Imp = m * ( v 1 - v 2 ) = m * Δ v
v 1 = - 10 i m/s,
v 2 = ( 20 * cos 40° ) i + ( 20 * sin 40° ) j =
= ( 20 * 0.766 ) i + ( 20 * 0.64278 ) j = ( 15.32 i + 12.855 j ) m/s
Δ v = ( 15.32 i + 12.855 j ) - ( - 10 i ) =
= 15.32 i + 12.855 j + 10 i = 25.32 i + 12.855 j
| Δv | = √ ( 25.32² + 12.855²) = √806.35 = 28.4 m/s
Imp = 0.2 kg * 28.4 m/s = 5.68 N-s
Answer: D ) 5.68 N-s.