Answer:
C. a dense region of positive charge existed somewhere in the atom.
Explanation:
Physicist Ernest Rutherford created the gold foil experiment in which he shot a beam of alpha particles at a sheet of gold foil, which then sent a few of the particles flying after they were deflected. Based on the information gathered after completing this experiment, Rutherford concluded that a dense region of positive charge existed somewhere in the atom.
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Answer:
Upper disk rotates at a constant angular velocity. The velocity at any height from stationery disk, say at x metres
where v is tangential velocity at radius r from the centre of disk
The radial component of velocity is given as
The z component of velocity is also given as
W=0
Total velocity,
Answer:
4.0 N
Explanation:
Sum the forces in the x direction:
∑F = ma
F − Fr = ma
Fr = F − ma
Fr = 5.00 N − (1.35 kg) (0.76 m/s²)
Fr = 4.0 N
Answer:
the displacement of the object is 5 units
Explanation:
The computation of the displacement of the object is shown below:
= Move to the right + move to the right - move to the left
= 6 units + 3 units - 4 units
= 9 units - 4 units
= 5 units
Hence, the displacement of the object is 5 units
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