Respuesta :
Answer:
increase in speed, Δv = 30.0m/s
Explanation:
Given:
Force on the spacecraft, F = 1.12 x 10⁻³ N,
Mass on spacecraft, m=290 kg
time, Δt = 3 month= 3 x 30 days =90 x 24 x 60x 60 seconds = 7776000 s
Now,
the acceleration (a) is given as:
a = F/m = (1.12 x 10⁻³)/290 = 3.86 x 10⁻⁶ m/s²
Applying Newton's equation of motion
v = u +aΔt
where,
v is the final velocity of the spacecraft after 3 months
u is the velocity of the spacecraft initially
v - u = (3.86 x 10⁻⁶) x 7776000
⇒ Δv = 30.0 m/s
The IKAROS spacecraft, whose mass is 290 kg and experiences a force of 1.12 mN, has a speed increase of 30 m/s after 3.0 months.
The spacecraft, with a mass (m) of 290 kg, experiences a net force (F) of 1.12 mN (1.12 × 10⁻³ N). We can calculate the acceleration (a) of the spacecraft using Newton's second law of motion.
[tex]F = m \times a\\\\a = \frac{F}{m} = \frac{1.12 \times 10^{-3}N }{290kg} = 3.86 \times 10^{-6} m/s^{2}[/tex]
Then, we will convert 3.0 months to seconds using the following conversion factors.
- 1 month = 30 days
- 1 day = 24 hours
- 1 hour = 3600 seconds
[tex]3.0 month \times \frac{30day}{1month} \times \frac{24h}{1day} \times \frac{3600s}{1 h} = 7.8 \times 10^{6} s[/tex]
The spacecraft experiences an acceleration of 3.86 × 10⁻⁶ m/s² for 7.8 × 10⁶ s. We can calculate the increase in the speed (Δv) using the definition of acceleration.
[tex]a = \frac{\Delta v }{t} \\\\\Delta v = a \times t = 3.86 \times 10^{-6} m/s^{2} \times 7.8 \times 10^{6} s = 30 m/s[/tex]
The IKAROS spacecraft, whose mass is 290 kg and experiences a force of 1.12 mN, has a speed increase of 30 m/s after 3.0 months.
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