A student woke up in the morning and checked the rain gauge in her backyard. The rain gauge was empty, but the grass was moist.
Which of the following best explains how this could happen?
A. It snowed, but did not rain during the night.
B. Run-off from a nearby river made the grass wet.
C. Water Vapor Condensed on the grass.
D. The water in the rain gauge evaporated, but the water on the grass did not.
Explain why you choose the answer you did.
THANK YOU!
A. It snowed, but did not rain during the night.
B. Run-off from a nearby river made the grass wet.
C. Water Vapor Condensed on the grass.
D. The water in the rain gauge evaporated, but the water on the grass did not.
Explain why you choose the answer you did.
THANK YOU!
2 answers:
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1) See graph in attachment
2) 10 s
3) 50 m
Explanation:
1)
In this problem, we have an object initially moving with a velocity of
v = 10 m/s
when the time is
t = 0 s
Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.
The result can be found in the graph in attachment.
Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to
which is equivalent to the gradient of the line in the velocity-time graph.
2)
In this part, we want to find after what time the body will stop its motion.
To do that, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
In this problem:
u = 10 m/s is the initial velocity of the body
is the acceleration
v = 0 m/s, because we want to find the time T at which the body will stop
Re-arranging the equation, we find:
3)
In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:
where
s is the distance covered
u is the initial velocity
t is the time
a is the acceleration
In this problem:
u = 10 m/s is the initial velocity
is the acceleration
t = 10 s is the time it takes for the body to stop (found in part 2)
Solving for s, we find the distance covered:
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
