A. A car moving at a constant speed on a flat, straight road. B. A vehicle traveling at a steady speed on a 10-degree incline. An object operates within an inertial reference frame if there is no net force acting upon it. According to Newton's second law, this implies that the object's acceleration also equals zero. Assessing the scenarios yields: A. A car moving at a constant speed on a flat road qualifies as an inertial reference frame, since its velocity and direction remain unchanged; thus, acceleration is zero. B. A car moving steadily up a 10-degree incline still constitutes an inertial reference frame, for similar reasons. C. A car accelerating after departing a stop sign does not represent an inertial frame due to its change in speed. D. A car driving at a steady speed around a curve cannot be considered an inertial reference frame since its direction is changing, resulting in a change in velocity and thus acceleration. Therefore, options A and B are correct.
Answer:
The outcome of adding 999mm to 100m is 101m.
Explanation:
That's my belief.
<span>a. To determine the velocity at which the camera strikes the ground:
v^2 = (v0)^2 + 2ay = 0 + 2ay
v = sqrt{ 2ay }
v = sqrt{ (2)(3.7 m/s^2)(239 m) }
v = 42 m/s
The camera impacts the ground with a speed of 42 m/s.
b. To calculate the duration it takes for the camera to reach the bottom:
y = (1/2) a t^2
t^2 = 2y / a
t = sqrt{ 2y / a }
t = sqrt{ (2)(239 m) / 3.7 m/s^2 }
t = 11.4 seconds
The camera descends for 11.4 seconds before hitting the ground.</span>
Answer:
The duration, t = 3.53 seconds
Explanation:
The following information is provided:
The equation to calculate the time t is expressed as:
...... (1)
Where
s denotes the distance in feet
We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet
Substituting the value of s into equation (1) yields:
t = 3.53 seconds
Thus, the time taken by the object is 3.53 seconds, which provides the required answer.
Answer:
Jari
Explanation:
To determine who is traveling faster, we need to evaluate their gradients. A steeper slope indicates a higher speed.
For Jari's path, starting point is (0, 0) and (6, 7) is another point.
The gradient is the difference in y divided by the difference in x:
Change in y=7-0=7
Change in x=6-0=6
Thus, the slope equals 7/6.
For Jade, her first point is (0, 10) and another is (6, 16).
Change in y=16-10=6
Change in x=6-0=6
Thus, the slope equals 6/6=1.
It's evident that 7/6 exceeds 6/6 or 1, proving Jari is quicker than Jade.
