Suppose that the acceleration of a model rocket is proportional to the difference between 160 ft/sec and the rocket's velocity.
1 answer:
Answer:
Explanation:
Given ,
dv / dt = k ( 160 - v )
dv / ( 160 - v ) = kdt
ln ( 160 - v ) = kt + c , where c is a constant
when t = 0 , v = 0
Putting the values , we have
c = ln 160
ln ( 160 - v ) = kt + ln 160
ln ( 160 - v / 160 ) = kt
(160 - v ) / 160 =
1 - v / 160 =
v / 160 = 1 -
v = 160 ( 1 - )
differentiating ,
dv / dt = - 160k
acceleration a = - 160k
given when t = 0 , a = 280
280 = - 160 k
k = - 175
a = - 160 x - 175
a = 28000
when a = 128 t = ?
128 = 28000
= .00457
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Answer: W = J
Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.
The work to transport an ion from a lower potential side to a higher potential side is calculated by
q is charge;
ΔV is the potential difference;
Potassium ion has +1 charge, which means:
p = C
To determine work in joules, potential has to be in Volts, so:
Then, work is
To move a potassium ion from the exterior to the interior of the cell, it is required J of energy.
Answer:
a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J
Explanation:
a) The inductance is a solenoid this given carrier
L =
The magnetic field inside the solenoid is
B = μ₀
hence the magnetic flux
Ф_B = B. A = μ₀
we substitute in the expression of inductance
L = N² μ₀ A /l
let's find the area of each turn
A = π r²
A = π 0.02²
A = 1.2566 10⁻³ m²
let's calculate
L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3
L = 3.29 10⁻⁴ H
b) The stored energy is
U = ½ L i²
let's calculate
U = ½ 3.29 10⁻⁴ 18²
U = 5.33 10⁻² J