At time t=0, a 2150kg rocket in outer space fires an engine that exerts an increasing force on it in the +x-direction. This force obeys the equation Fx=At2, where 1.45s is time, and has a magnitude of 781.25N when t = 1.45s .

Find the SI value of the constant A.

Rearrange the equation:

A = F_x / t^2
Then simply substitute the numbers:
.: A = (781.25/1.45^2) N/s^2

Answer in three significant figures.

To find the SI value of the constant A in the equation Fx = At^2, we need to use the given information about the magnitude of the force (781.25N) when t = 1.45s.

First, let's substitute the given values into the equation:

781.25N = A(1.45s)^2

Next, we can solve for A by isolating it on one side of the equation. Let's divide both sides of the equation by the square of 1.45s:

A = 781.25N / (1.45s)^2

Calculating the right side of the equation:

A = 781.25N / (1.45s * 1.45s)

A = 781.25N / 2.1025s^2

A ≈ 371.93 N/s^2

Therefore, the SI value of the constant A is approximately 371.93 N/s^2.

To find the SI value of the constant A, we can use the given information about the force and time.

Given:
Mass of rocket (m) = 2150 kg
Force in the x-direction (Fx) = 781.25 N
Time (t) = 1.45 s

The force exerted by the engine is given by the equation Fx = At^2.

Plugging in the given values, we have:
781.25 N = A * (1.45 s)^2

Simplifying the equation, we get:
781.25 N = A * (2.1025 s^2)

Next, we isolate the constant A by dividing both sides of the equation by (2.1025 s^2):
A = 781.25 N / (2.1025 s^2)

Simplifying further, we calculate the value of A:
A = 372.02 N/s^2

Therefore, the SI value of the constant A is 372.02 N/s^2.