Answer:
Explanation:
The electric field inside a parallel plate capacitor is
where A is the area of one of the plates, and Q is the charge on the capacitor.
The electric force on the electron is
where q is the charge of the electron.
By definition the capacitance of the capacitor is given by
Plugging this identity into the force equation above gives
The work done by this force is equal to change in kinetic energy.
W = Fx = (30q)(0.05) = 1.5q = K
The charge of the electron is 
Therefore, the kinetic energy is 
Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:
where
A) f = 1.8 rev/s = 2 Hz
<span>T = 1 / f = 0.55s
B) not really sure..srry
C) </span><span>T = 2 pi √ ( L / g ) </span>
<span>0.57 = 2 x 3.14 x √ ( 0.2 / g )
</span><span>
g = 25.5 m/s²
</span>
Hope this helps a little at least.. :)
Answer: B
Explanation:
Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.
Answer:
-0.01 mm
Explanation:
We are given that
The value of one division of vernier scale =0.5 mm
The value of one main scale division=0.49 mm
We have to find the value of least count of the instrument in mm.
We know that
Leas count of vernier caliper=1 main scale division-1 vernier scale division
Least count of vernier caliper=0.49-0.50=-0.01 mm
Hence, the least count of the instrument=-0.01 mm
Answer: -0.01 mm
