Answer:
σ₁ = 
C/m²
σ₂ = 
C/m²
Explanation:
Provided Information:
i) Smaller sphere's radius ( r ) = 5 cm.
ii) Larger sphere's radius ( R ) = 12 cm.
iii) Electric field at the larger sphere's surface ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m
Charge (Q₁) = 572.8 
C
Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.
V = constant
∴
= 
C
Surface charge density ( σ₁ ) for the larger sphere.
Calculated Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².
σ₁ = 
= 
= C/m².
Surface charge density ( σ₂ ) for the smaller sphere.
Calculated Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² = 0.0314 m².
σ₂ =
= 
= C/m²
through the Doppler effect. The formula for apparent frequency is derived as F apparent = F real x (Vair ± Vobserver) / (Vair ± Vsource). In this scenario, should the observer move towards the source—place a positive sign in the numerator and a negative in the denominator. Since the observer approaches the wall, we apply the formula to derive the necessary speed.
The following values have been provided:
weight w = 240 lb = 1,067.52 N
energy E = 3,000 J
The equation for potential energy is:
E = w h
where h indicates the height that the person needs to ascend, therefore:
h = 3000 / 1067.52
h = 2.81 m
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</span>
<span>Thus, he must ascend 2.81 meters</span>
Answer:3.87*10^-4
Explanation:
To determine the mass reduction, delta mass Xe, of the xenon nucleus due to its decay, we first use the provided wavelength of the gamma radiation to calculate its frequency via c = freq*wavelength.
From C=f*lambda we set up: 3*10^8=f*3.44*10^-12.
Solving gives frequency F=0.87*10^20 Hz.
Next, we calculate the emitted energy using the equation E=hf, which translates to E=f*Planck's constant.
Thus, E=0.87*10^20*6.62*10^-34, resulting in E=575.94*10^(-16).
This energy is then converted from joules to MeV.
Utilizing the formula E=mc^2, with c^2 = 931.5 MeV/u, enables us to find the reduction in mass, yielding
3.87*10^-4 u.
