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Complete question is;
A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?
Answer:
v = 12 m/s
Explanation:
We are given;
Initial velocity; u = 0 m/s (because ship starts from rest)
Acceleration; a = 4 m/s²
Time; t = 3 s
To find velocity after 3 s, we will use Newton's first equation of motion;
v = u + at
v = 0 + (4 × 3)
v = 12 m/s
%d is a format specifier that is a placeholder for an int value. It tells the compiler that we want to print an integer value that is present in variable a. In this way there are several format specifiers in c.
Answer:
2.572 m/s²
Explanation:
Convert the given initial velocity and final velocity rates to m/s:
- 65 km/h → 18.0556 m/s
- 35 km/h → 9.72222 m/s
The motorboat's displacement is 45 m during this time.
We are trying to find the acceleration of the boat.
We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.
Substitute the known values into the equation.
- (9.72222)² = (18.0556)² + 2a(45)
- 94.52156173 = 326.0046914 + 90a
- -231.4831296 = 90a
- a = -2.572
The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².
