Answer:
because they are found freely in nature uncombined so they are highly reactive with other elements
Answer:
t ’=
, v_r = 1 m/s t ’= 547.19 s
Explanation:
This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.
By the time the boat goes up the river
v_b - v_r = d / t
By the time the boat goes down the river
v_b + v_r = d '/ t'
let's subtract the equations
2 v_r = d ’/ t’ - d / t
d ’/ t’ = 2v_r + d / t
In the exercise they tell us
d = 1.22 +1.45 = 2.67 km= 2.67 10³ m
d ’= 1.45 km= 1.45 1.³ m
at time t = 69.1 min (60 s / 1min) = 4146 s
the speed of river is v_r
t ’=
t ’=
In order to complete the calculation, we must assume a river speed
v_r = 1 m / s
let's calculate
t ’=
t ’= 547.19 s
<span>PV = nRT
moles of H2 = 1/2 = 0.5
moles of He = 1/4 =0.25
T = 273 + 27
partial pressure of H2
Px1 = 0.5x0.083x300
P=12.45 atmospheres
PP of He
px1 = 0.25x0.083x300
P =6 22 atmospheres
Totla pressure = 6.22 + 12.45 = 18.68 atm</span>
Answer: C) 0.25 m
Explanation: In order to explain this problem we have to consider the Faraday law, we have:
ε=-dФ/dt where ε the emf induced by the change of the magnetic field given by dФ/dt.
then ε=I*R=17*6=102 V
We have a coil then we have the magnetic flux as follow:
Ф=N*A*B then we have
dФ/dt= N*A*dB/dt where A and N is the area and number of turn of the coil.
A=π*R^2 where R is the radius of teh coil.
Finally we have;
dФ/dt= N*π*R^2*dB/dt then
R= [dФ/dt/(N*π*dB/dt)]^1/2= [102/(180*π*3)]1/2=245.2*10^-3=≅0.25m