Answer:
per mole of pentane = 3157.53 kJ/mol
Explanation:
Given:
Mass of pentane, m = 0.468 gram
Molar mass of pentane, M = 72.15
Now, mol of pentane, n = mass/M = 0.468/72.15 = 0.00648 mol of C5H12
Now,
ΔT = 23.65 - 20.45 = 3.2°C
Heat capacity of the calorimeter, C = 2.21 kJ/°C
Specific heat capacity of the water, Cp = 4.184 J/g.°C
Now,
the heat gained = the heat lost

also,


or

and
Now,

we have,
(Here negative sign depicts the release of the heat)
per mole of pentane =-20460.8/(0.00648 ) = 3157.53 kJ/mol
Rw^2 = GmM/r^2
<span> Leads to
</span><span> w^2 r^3 = GM
</span><span> (2pi /T) ^2 r^3 = GM
</span><span> 4pi^2 r^3 = GM T^2
</span><span> r^3 = GM T^2 / 4pi^2
</span><span> Work out r^3 then r.
</span> T = 125 min = 125(60) = 7500 s
<span> R = 6.38E6 m
</span><span> m = 5.97E24 kg
</span><span> G = 6.673E-11
</span> r=<span>
8279791.78</span><span> m
Since r = radius R of Earth + height above urface,h
</span><span> h = r - R = </span><span>
8279791.78 - </span>6.38E6 = <span>
<span>1899791.78 m
h=</span></span><span>
<span>1899.79178 Km</span></span>
Answer:
<h2>9.375Nm</h2>
Explanation:
The formula for calculating torque τ = Frsin∅ where;
F = applied force (in newton)
r = radius (in metres)
∅ = angle that the force made with the bar.
Given F= 25N, r = 0.75m and ∅ = 30°
torque on the bar τ = 25*0.75*sin30°
τ = 25*0.75*0.5
τ = 9.375Nm
The torque on the bar is 9.375Nm
Answer:
53.63 μA
Explanation:
radius of solenoid, r = 6 cm
Area of solenoid = 3.14 x 6 x 6 = 113.04 cm^2 = 0.0113 m^2
n = 17 turns / cm = 1700 /m
di / dt = 5 A/s
The magnetic field due to the solenoid is given by
B = μ0 n i
dB / dt = μ0 n di / dt
The rate of change in magnetic flux linked with the solenoid =
Area of coil x dB/dt
= 3.14 x 8 x 8 x 10^-4 x μ0 n di / dt
= 3.14 x 64 x 10^-4 x 4 x 3.14 x 10^-7 x 1700 x 5 = 2.145 x 10^-4
The induced emf is given by the rate of change in magnetic flux linked with the coil.
e = 2.145 x 10^-4 V
i = e / R = 2.145 x 10^-4 / 4 = 5.36 x 10^-5 A = 53.63 μA