Answer:
1. Golf ball is in the range of only particulate detectable range
2, alpha partcle is in ultraviolet range of wavelength
3.bullet is in xray range of wavelength
All in the EMW spectrum
If<span> The </span>Sun<span> Went Out, How Long </span>Could<span> Life On </span>Earth<span> Survive? ... (which is actually physically impossible), the </span>Earth would stay<span> warm—at least ... from the planet's core </span>would<span> equal the</span>heat<span> that the </span>Earth<span> radiates into space, ... Photosynthesis </span>would<span> halt immediately, and </span>most<span> plants</span>would<span> die </span>in<span> a few </span>weeks<span>.</span>
Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
