Answer:
6.77 minutes
Explanation:
172 degree - 78 degree = (185 degree - 78 degree)e−2 k
=> 94 = 107
e−2 k => 94 ÷ 107
k => ln (94÷107) / 2
147 - 78 = (185 - 78)e ^[ln (94÷107) / 2]
=> 69 = 107 e^ [ln (94÷107) / 2]
e^[ln (94÷107) / 2] =69 ÷ 107
=> t = [ln (69 ÷ 107)] ÷ [ln (94÷107) / 2]
t=> -0.4387 ÷ -0.0648
t => 6.77 minutes.
Therefore, the final answer to the question is 6.77 minutes.
Answer:
The travel would take 6.7 years.
Explanation:
The equation for an object moving in a straight line with acceleration is:
x = x0 + v0 t + 1/2a*t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
In a movement with constant speed, a = 0 and the equation for the position will be:
x = x0 + v t
where v = velocity
Let´s calculate the position from the Earth after half a year moving with an acceleration of 1.3 g = 1.3 * 9.8 m/s² = 12.74 m/s²:
Seconds in half a year:
1/2 year = 1.58 x 10⁷ s
x = 0 m + 0 m/s + 1/2 * 12.74 m/s² * (1.58 x 10⁷ s)² = 1.59 x 10¹⁵ m
Now let´s see how much time it takes the travel to the nearest star after this half year.
The velocity will be the final velocity achived after the half-year travel with an acceleration of 12.74 m/s²
v = v0 + a t
Since the spacecraft starts from rest, v0 = 0
v = 12.74 m/s² * 1.58 x 10⁷ s = 2.01 x 10 ⁸ m/s
Using the equation for position:
x = x0 + v t
4.1 x 10¹⁶ m = 1.59 x 10¹⁵ m + 2.01 x 10 ⁸ m/s * t
(4.1 x 10¹⁶ m - 1.59 x 10¹⁵ m) / 2.01 x 10 ⁸ m/s = t
t = 2.0 x 10⁸ s * 1 year / 3.2 x 10 ⁷ s = 6.2 years.
The travel to the nearest star would take 6.2 years + 0.5 years = 6.7 years.
Answer:
conduction
Explanation:
Conduction is a method of heat transfer, where two objects exchange heat by touching, but does not contribute to moving heat through earth's atmosphere, because the atmosphere is so sparse it does not transmit heat through conduction effectively. Insulation is not a method of heat transfer.
