Answer:
A line crossing the x-axis in a velocity-time graph means that the moving object has changed its direction.
Explanation:
Velocity-Time graph:
A velocity-time graph is a two dimensional graph with velocity at its y-axis and time at its x-axis. At any point, value of y represents the velocity and value of x represents the time. The slope of the graph gives us the acceleration or deceleration of the moving object.
In a velocity-time graph:
- A straight line represents constant velocity.
- A diagonal line means that the velocity of a body is changing.
*Referring to the figure attached with the answer*
The velocity of the moving object increases at a constant rate for the first 10 minutes. Then the velocity is 60 m/min for the next 5 minutes. After that the velocity is decreasing. Till 30th minute when the velocity is at 0 m/min.
What happens here?
Velocity is a vector quantity. It has some direction. In a velocity-time graph, we are only concerned with two directions of velocity:
- Forward direction
- Backward direction
So, the object stops at 30th minute and starts moving in the reverse direction after that with an increasing velocity. <u>The point where the line cuts the x-axis is basically the point where the object starts moving in the reverse direction.</u>
the answer is c) when it becomes excessive, uncontrollable , and affect daily life
Answer:
A flat, horizontal line
Explanation:
A flat, horizontal line indicates a phase change.
The temperature does not increase because the added heat goes into converting one phase into another.
A is wrong. A downward-sloping line indicates that the temperature is decreasing with time.
C is wrong. An upward-sloping line indicates that the temperature is increasing with time.
To solve this problem, we can use the cosine formula for
calculating the length of the displacement:
c^2 = a^2 + b^2 – 2 a b cos θ
where c is the displacement, a = 3.5 km, b = 4.5 km, and θ
is the angle inside the triangle
Since the geeze turned 40° from west to north, so the
angle inside the triangle must be:
θ = 180 – 40 = 140°
c^2 = 3.5^2 + 4.5^2 – 2 (3.5) (4.5) cos 140
c^2 = 56.63
c = 7.53 km
<span>So the magnitude of the displacement is 7.53 km</span>
Answer:
384.6 m
Explanation:
The length of the runway airport should be long enough to accommodate the aircraft during its acceleration from rest to 161 km/h at rate of 2.6 m/s. We can use the following equation of motion to solve for this:
where v0 = 0 m/s is the initial velocity of the airplane when it start accelerating, v = 161 km/h = 161*1000*(1/60)(1/60) = 44.72 m/s is the airborn speed, a = 2.6 m/s2 is the acceleration, and 
is the distance of the runway, which we care looking for
