Answer:
Explanation:
i )
When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2
So charged stored in it will remain unchanged .
ii )
Potential difference = charge / capacitance
in the first case potential difference = Q / C
in the second case potential difference = Q / 2C
So potential difference becomes half .
iii ) electric field = potential diff / plate separation
in the first case electric field = Q / (d x C )
in the second case electric field = 2 Q / (d x 2C)
= Q / (d x C )
So electric field remains unchanged .
iv)
energy stored in first case = Q² / 2C
In the second case energy stored = Q² / 2x2C
so energy stored becomes half .
Answer: im not gonna give i to you just do 15+15=_+ 5.6+6.4 easy
Explanation: i took the test and got a 100%
Answer:
A student is conducting a pendulum experiment. Which of the following pieces of safety equipment would be the most vital to conduct this test?
Explanation:nduebidndo eyn h ehj jd
jknsjk Jjkfnjkjnuifvr
The final temperature of the system is 32.5°
we know, H = mcT
where, H = Heat content of the body
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry
The net heat remains constant i.e.
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x
x (T - 10)
⇒ 14,850 - 330T = 183.21T - 1832
⇒ - 513.21 T = - 16682
or T = 32.5°
Answer:
width of slit(a)≅ 0.1mm
Explanation:
Wave length of laser pointer =λ = 685 nm
Distance between screen and slit = L = 5.5 m
Width of bright band = W=8.0cm=0.08m
width of slit=a
recall the formula;
W=(2λL)/a
a=2λL/W
a=(2 *685*10⁻⁹*5.5m)/0.08m
a=7535*10⁻⁹/0.08
a=94187.5 *10⁻⁹
a=0.0000941875m
a=0.0941875mm
a≅0.1mm
