A spaceship starting from a resting position accelerates at a constant rate of 9.8 m/s. How far will the spaceship travel if its
1 answer:
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Answer:
Explanation:
Hello,
In this case, considering that the acceleration is computed as follows:
Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:
Regards.
Answer:
8.94*10^22 kg
Explanation:
Given that
Mass of Lo, M = ?
Radius of Lo, r = 1.82*10^6 m
Acceleration on Lo, g = 1.80 m/s²
Gravitational constant, G = 6.67*10^-11
Using the formula
g = GM/r²
Solution is attached below
Answer is 8.94*10^22 kg
Answer:
The final speed of the crate is 12.07 m/s.
Explanation:
For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:
Now, we can calculate the final speed of the crate at the end of 10.0 m:
For the next 10.5 meters we have frictional force:
So, the acceleration is:
The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:
Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.
I hope it helps you!