Response:
The car's acceleration magnitude is 35.53 m/s²
Details:
Given;
acceleration of the truck, 
= 12.7 m/s²
mass of the truck, 
= 2490 kg
mass of the car, 
= 890 kg
let the acceleration of the car during the collision = 
Using Newton's third law of motion;
The force exerted by the truck equals the force exerted by the car.
The car's force acts in the opposite direction.
Thus, the car's acceleration magnitude is 35.53 m/s²
Response:
83%
Clarification:
At the surface, the weight can be expressed as:
W = GMm / R²
where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.
When in orbit, the weight is given by:
w = GMm / (R+h)²
where h indicates the shuttle's altitude above Earth's surface.
The weight ratio is as follows:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
For R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
Thus, the shuttle maintains 83% of its weight as it orbits.
The height is h = 17 10⁶ meters above the surface of Mars. To determine this, we apply Newton's second law according to the universal law of gravitation, represented by F = m a. The centripetal acceleration a is expressed as v² / r. Applying the gravitational force we have G m M / r² = m v² / r. Given that the speed of the object remains constant, we derive v from d / t, where d is the circumference and t is the orbital period. Substituting gives us d = 2π r and v = 2π r / T. Replacing these values leads to the equation G M / r² = (4π² r² / T) / r, so r³ = G M T² / 4π². Converting time into SI units, T = 24.66 h converts to 88776 seconds. Ultimately, the computed value of r is 2,045 10⁶ m, and after subtracting Mars’ radius of 3.39 10⁶ m, we find the height h to be 17 10⁶ m.
