Answer:
The relationship between amperage, voltage, and power is that power equals the amperage quantity times the amount of voltage.
Explanation: brainliest pls
1,000 W = 1 kW
100 W = 0.1 kW
(0.1 kW) x (6 h) = 0.6 kWh <=== energy
(0.6 kWh) x (£0.1359/kWh) = £0.0815 <=== cost of it
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
Answer:
the coefficient of volume expansion of the glass is 
Explanation:
Given that:
Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³
temperature of the glass flask and mercury= 1.00° C
After heat is applied ; the final temperature = 52.00° C
Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C
Volume of the mercury overflow = 8.50 cm^3 = 8.50 × 10⁻⁶ m³
the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K
The increase in the volume of the mercury = 10⁻³ m³ × 51.00 × 1.80 × 10⁻⁴
The increase in the volume of the mercury = 
Increase in volume of the glass = 10⁻³ × 51.00 × 
Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask
the mercury overflow = 
Thus; the coefficient of volume expansion of the glass is
