Answer:
Please find the answer in the explanation
Explanation:
When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above
The slope of any time graph can not give you distance or displacement except for position - time graph.
When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.
You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.
The velocity of the stone when it is 5.25 m above the ground is determined as 6.93 m/s.
The velocity of the stone at the given displacement is calculated as follows;
K.E = ΔP.E
¹/₂mv² = mgh
v = √2gΔh
v = √[2(9.8)(7.7 - 5.25)]
v = 6.93 m/s
Thus, the velocity of the stone when it is 5.25 m above the ground is determined as 6.93 m/s.
Learn more about velocity here: brainly.com/question/4931057
#SPJ1
Answer:
190 is the average velocity.
Explanation:
Just add them together
a. the volume of the container that the gas is in
b. the percentage of the gas
c. the percentage of the other gases that comprise the atmosphere
d. the total pressure in the atmosphere
Answer:
C.)
Explanation:
i said c because. idk i guessed sry
At the peak of its trajectory, the ball's vertical velocity (v1,y) is zero, whereas its horizontal velocity (v1,x) maintains its consistency. The accelerations a1,x and a1,y are 0 and -9.8 m/s² respectively.
The physical concept analyzed here is projectile motion. Considering the trajectory of a ball, we can separate its motion into horizontal and vertical components. At the peak of its trajectory (time t1), the ball's vertical velocity (v1,y) is zero, because it temporarily stops moving upwards before it begins to fall again. However, since there is no acceleration in the horizontal direction (a1,x), the ball keeps moving horizontally with constant velocity (v1,x).
The acceleration in the vertical direction (a1,y) is still the acceleration due to gravity (-9.8 m/s²). This is because even at the peak of the trajectory, the ball is still accelerating downwards. Therefore, the correct answer should include v1,y = 0 m/s, v1,x = consistent value, a1,x = 0 m/s², and a1,y = -9.8 m/s². So the most probable answer is D) 15.0, 0, 0, -9.80, assuming the ball's horizontal velocity is 15 m/s. Please check the exact value of horizontal velocity (v1,x) in the settings of your problem.
#SPJ12