91Ó°ÊÓ

To find the acceleration needed for the spacecraft, we use the formula for acceleration: \( a = \frac{\Delta v}{\Delta t} \). Let's break that down:- \( \Delta v \) represents the change in velocity, which is from rest (0) to \( 0.10c \), where \( c \) is the speed of light, or \( 3.0 \times 10^8 \text{ m/s} \). Hence, \( \Delta v = 0.10 \times 3.0 \times 10^8 \text{ m/s} \).- We need to convert the given time of 3 days into seconds to align with the SI unit system. So, \( 3 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} = 259,200 \text{ seconds} \).Plug these values into the acceleration formula to get:\[a = \frac{0.10 \times 3.0 \times 10^8}{259,200} \approx 1156 \text{ m/s}^2\]This is the constant acceleration required for the spacecraft to reach 0.10 times the speed of light in 3.0 days.

In physics, comparing acceleration with Earth's gravity, \( g \), is common because it provides an intuitive grasp of the scale of acceleration. Earth’s gravitational acceleration, \( g \), is approximately \( 9.81 \text{ m/s}^2 \).
To convert the calculated acceleration to \( g \) units, use the formula:\[a_{g} = \frac{a}{9.81}\]- Substitute the value of \( a \) from the previous calculation: \( a = 1156 \text{ m/s}^2 \).
- So, \( a_g = \frac{1156}{9.81} \approx 117.87g \).Therefore, the constant acceleration required is about \( 117.87 \g \) when expressed in Earth gravity units.

To calculate the force required to achieve this acceleration, we apply Newton's second law of motion:
\( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
- Here, \( m = 1.20 \times 10^6 \text{ kg} \) (the mass of the spacecraft).- And \( a = 1156 \text{ m/s}^2 \).Substituting these values into the formula:\[F = 1.20 \times 10^6 \times 1156 \approx 1.39 \times 10^9 \text{ N}\]This calculation shows that a force of approximately \( 1.39 \times 10^9 \) newtons is needed to propel the spacecraft to the desired speed within the given timeframe.

Understanding the concept of a light-month can provide insight into the vastness of space distances. 1 light-month is the distance that light travels in one month.- Light speed, \( c = 3.0 \times 10^8 \text{ m/s} \).- Number of seconds in a month: \( 30 \times 24 \times 3600 \text{ s} \).To find the distance light travels in 5 months:\[\text{Distance} = 5 \times c \times \text{seconds in a month}\]Plug in the values:\[\text{Distance} = 5 \times 3.0 \times 10^8 \times 30 \times 24 \times 3600 \approx 3.89 \times 10^{15} \text{ m}\]The next step is calculating the travel time for this distance at a constant speed of \( 0.10c \).- The speed of the ship, \( v = 0.10 \times 3.0 \times 10^8 \text{ m/s} \).To find the time \( t \), use: \( t = \frac{d}{v} \)\[t = \frac{3.89 \times 10^{15}}{0.10 \times 3.0 \times 10^8} \approx 1.30 \times 10^7 \text{ s}\]Converting seconds to different units, such as hours or days, can be done for practical purposes, knowing there are 3600 seconds in an hour.