Answer:
a I think hope this helps
consider the motion in x-direction
= initial velocity in x-direction = ?
X = horizontal distance traveled = 100 m
= acceleration along x-direction = 0 m/s²
t = time of travel = 4.60 sec
Using the equation
X =
t + (0.5)
t²
100 =
(4.60)
= 21.7 m/s
consider the motion along y-direction
= initial velocity in y-direction = ?
Y = vertical displacement = 0 m
= acceleration along x-direction = - 9.8 m/s²
t = time of travel = 4.60 sec
Using the equation
Y =
t + (0.5)
t²
0 =
(4.60) + (0.5) (- 9.8) (4.60)²
= 22.54 m/s
initial velocity is given as
= sqrt((
)² + (
)²)
= sqrt((21.7)² + (22.54)²) = 31.3 m/s
direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg
The answer is C as there is more force on the left side ( excess of 5 N) which therefore pushed it to the right with a force of 5 N!
See the graph in attachment
Explanation:
In this problem we have to draw a velocity-time graph for an object travelling initially at -3 m/s, then slowing down and turning around.
In the graph, we see that the initial velocity at time t = 0 is

and it is negative, so below the x-axis.
Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).
After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.
Learn more about velocity:
brainly.com/question/5248528
#LearnwithBrainly
Answer:
I gotchu fam. 0.0435 kg
Explanation:
m1v1+m2v2=m1v1f+m2v2f
(0.5)(23)+(m2)(0)=(0.5)(17)+(m2)(69)
11.5=8.5+69m2
3=69m2
0.0435 kg