The answer is 14000 millimeters
Answer:
(a) Total north component = (26.89 + 25) = 51.89 metres
(b) Total west component = (-34.83 + 52) = 17.17 metres.
Explanation:
Do northwest components first.
North component = (sin 33.9 x 44) =26.89 metres.
West component = (cos 33.9 x 44) = -34.83metres.
Total west component = (-34.83 + 52) = 17.17 metres.
Total north component = (26.89 + 25) = 51.89 metres
Answer:
16J
Explanation:
From hookes law
The work done in a spring is given as W =1/2ke^2
Given that the force constant (k) is constant in the spring material
We have that 2W = e^2
Let W1 = 4J e1= 2cm e2 = 4cm
Let W2 be the work required to stretch it an additional 4cm
W1/ W2 = e1^2/e2^2
W2 = W1* e2^2 / e1^2
= 4* 4^2 /2^2
=4× 16 / 4
= 16J
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.