Answer:
i) C decreases
ii) Q remains constant
iii) E remains constant
iv) ΔV increases
Explanation:
i)
We know, capacitance is given by:


<em>In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.</em>
ii)
As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.
iii)
Electric field between the plates is given as:

where:
charge density, 
<em>As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.</em>
iv)
- From the basic definition of voltage we know that it is the work done per unit charge to move it through a distance.
- Here we increase the distance so the work done per unit charge increases.
The answer is
2.5 N
B
this needs to be 20 letters long so this part doesn't matter
Answer:
Explanation:
Given
Initial speed 
distance traveled before coming to rest 
using equation of motion

where v=final velocity
u=initial velocity
a=acceleration
s=displacement

for 
using same relation we get

divide 1 and 2 we get


So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed
No but the sun could be a white dwarf stellar remnant.
Answer:
102597.6 Pa
Explanation:
mass, m = 1.25 g
Force, F = m x g = 1.25 x 9.8 x 10^-3 = 0.01225 N
radius, r = 0.195 mm = 0.195 x 10^-3 m
Area, A = πr² = 3.14 x 0.195 x 0.195 x 10^-6 m^2
A = 1.19 x 10^-7 m^2
Pressure is defined as the thrust acting per unit area.
P = Force / Area
P = 0.01225 N / (1.10 x 10^-7)
P = 102597.6 Pa
Thus, the pressure exerted is 102597.6 Pa.