Answer:
a) R(fw) = 46.75*10⁶ (N)
b) R(rwi) = 70.125*10⁶ [N]
Explanation: See Attached file (the rectangle stands for a jet)
The diagram shows forces acting on the jet
Let R(fw) Reaction of front wheels and
R(rw) Reaction of rear wheels
Now we apply the Stevin relation, for R(fw) and a jet weight as follows
R(fw)/ 4 = 187*10⁶ / 16
Then :
R(fw) = ( 1/4) *187*10⁶ ⇒ R(fw) = 46.75*10⁶ (N)
And e do the same for the reaction on rear wheels
R(rw) / 12 = 187*10⁶ /16 ⇒ R(rw) =(3/4)*187*10⁶
R(rw) = 140,25*10⁶ [N]
The last expression is for the whole reaction, and must be devide by 2
because that force is exerted for two wheels, therefore on each of the two rear wheels the reaction will be:
R(rwi) = 70.125*10⁶ [N]
Answer:
the answer is 1, oil and water
Answer:
no, 3 porces is more tha 2 so the power between the 3 should be more than 2
Explanation:
Answer:
Explanation:
Inductance L = 1.4 x 10⁻³ H
Capacitance C = 1 x 10⁻⁶ F
a )
current I = 14 .0 t
dI / dt = 14
voltage across inductor
= L dI / dt
= 1.4 x 10⁻³ x 14
= 19.6 x 10⁻³ V
= 19.6 mV
It does not depend upon time because it is constant at 19.6 mV.
b )
Voltage across capacitor
V = ∫ dq / C
= 1 / C ∫ I dt
= 1 / C ∫ 14 t dt
1 / C x 14 t² / 2
= 7 t² / C
= 7 t² / 1 x 10⁻⁶
c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance
energy stored in inductor
= 1/2 L I²
energy stored in capacitor
= 1/2 CV²
After time t
1/2 L I² = 1/2 CV²
L I² = CV²
L x ( 14 t )² = C x ( 7 t² / C )²
L x 196 t² = 49 t⁴ / C
t² = CL x 196 / 49
t = 74.8 μ s
After 74.8 μ s energy stored in capacitor exceeds that of inductor.
