Answer:
66.24 Volts
Explanation:
W = Amount of work done in moving the charge from negative to positive terminal = 2.186 J
Q = Amount of charge being moved from negative to positive charge = 0.033 C
ΔV = EMF of the battery
Amount of work done in moving the charge from negative to positive terminal is given as
W = Q ΔV
2.186 = (0.033) ΔV
ΔV = 66.24 Volts
Answer:
warmer
Explanation:
I really think it just gets hot
Answer:
Part a)
Part b)
Explanation:
As we know that capacitor plate is connected across 2.1 V and after charging it is disconnected from the battery
So here we can say that charge on the plates will remain conserved
So we will have
now dielectric is removed between the plates of capacitor
so new potential difference between the plates
Part b)
Now the capacitor plates are again isolated and unknown dielectric is inserted between the plates
So again charge is same so potential difference is given as
Answer:
Explanation:
Given the height reached by a balloon after t sec modeled by the equation
h=1/2t²+1/2t
a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t
If h(t)=1/2t²+1/2t
h(40) = 1/2(40)²+1/2 (40)
h(40) = 1600/2 + 40/2
h(40) = 800 + 20
h(40) = 820 feet
The height of the balloon after 40 secs is 820 feet
b) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
when v = 0sec
v(0) = 0 + 1/2
v(0) = 1/2 ft/sec
at v = 30secs
v(30) = 30 + 1/2
v(30) = 30 1/2 ft/sec
average velocity = v(30) - v(0)
average velocity = 30 1/2 - 1/2
average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec
c) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
The velocity of the balloon after 30secs will be;
v(30) = 30+1/2
v(30) = 30.5ft/sec
The velocity of the balloon after 30 secs is 30.5 feet/sec
