In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s
1 answer:
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Velocity vs. time graph shows the acceleration as a slope whereas displacement vs. time graph shows the velocity as a slope. So, the given statement is false.
Answer: Option B
<u>Explanation:</u>
To understand the acceleration graphically, consider the x axis of the graph as the run and the y axis as the velocity rise. Now, as we all know that,
We can estimate this through the graph. let's draw the motion of an object with time if it's velocity is changing in every second by 4 m/s. Now if we draw this on graph, we will see that there is a slope between the two corresponding values of time and velocity. This slope defines the acceleration for the object with time.
Now, in the same way, if we draw a distance and time graph respective to the y and x axis; we'll get a slope which defines the velocity of the object i.e. change in distance with time.
Hence, with a velocity vs time graph, we get the slope for acceleration whereas with the distance and time graph, we get the slope for velocity. So both the cases, we see there is no velocity slope on an acceleration and time graph. Hence the statement is false.
This question is incomplete, the complete question is;
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.
(a) Determine the forces and and the couple
(b) Determine the sum of the moments about the right end of the beam.
(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?
Answer:
a)
the x-component of the force at A is = 0
the y-component of the force at A is = 400 N
the couple acting at A is; = 146 N-m
b)
the sum of the momentum about the right end of the beam is; ∑ = 0
c)
the equivalent force acting at the left end is; F = -400J ( N)
the couple acting at the left end is; M = - 146 N-m
Explanation:
Given that;
The sum of the forces acting on the beam is zero ∑f = 0
Sum of the moments about the left end of the beam is also zero ∑ = 0
Vector force acting at A, =
+
Now, From the image, we have;
a)
∑f = 0
- 600j + 200j = 0i + 0j
+
- 600j + 200j = 0i + 0j
+ (
- 400)j = 0i + 0j
now by equating i- coefficients'
= 0
so, the x-component of the force at A is = 0
also by equating j-coefficient
- 400 = 0
= 400 N
hence, the y-component of the force at A is = 400 N
we also have;
∑ = 0
- ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0
- 30 N-m - 228 N-m + 112 Nm = 0
- 146 N-m = 0
= 146 N-m
Therefore, the couple acting at A is; = 146 N-m
b)
The sum of the moments about right end of the beam is;
∑ = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)(
) +
∑ = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m
∑ = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m
∑ = 0
Therefore, the sum of the momentum about the right end of the beam is; ∑ = 0
c)
The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.
The equivalent force at the left end will be;
F = -600j + 200j (N)
F = -400J ( N)
Therefore, the equivalent force acting at the left end is; F = -400J ( N)
Also couple acting at the left end
M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)
M = -(30 N-m) + (112 N-m) - ( 228 N-m))
M = 112 N-m - 258 N-m
M = - 146 N-m
Therefore, the couple acting at the left end is; M = - 146 N-m
Answer:
T = 4.905[N]
Explanation:
In order to solve this problem we must perform a sum of forces on the vertical axis.
∑Fy = 0
We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.