Answer: A a decrease of 8000
Explanation: 10,500 - 2,500= 8000
1. A. 6.00 sec
The graph shows the velocity of an object (y-axis) versus the time (x-axis). In order to find when the magnitude of the velocity reaches 36.00 km/h, we should find the time t (x-coordinate) at which the velocity (y-coordinate) is 36.
By looking at the graph, we see that this occurs when t=6.00 s.
2. A. positive acceleration
In a velocity-time graph like this one, the slope of the curve corresponds to the acceleration of the object. In fact, acceleration is defined as:
where 
is the variation of velocity and 
is the variation of time. We see that this quantity corresponds to the slope of the curve in the graph (in fact, represents the increment of the y coordinate, while represents the increment of the x coordinate). So, a positive slope means a positive acceleration: in this case, the slope is positive, so the acceleration is also positive.
Answer:
7.0s
Explanation:
Mass = 0.41kg
F= 81N
t = 0.22s
¤ = 29°
Lo = 86m
From impulse equation,
F*t = m* v
81 * 0.22 = 0.41 * v
Vo = 17.82 / 0.41
Vo = 43.46m/s
Vx= velocity across horizontal plane
Vy = velocity across vertical plane
Vx = Vo * cos ¤
Vy = Vo * sin ¤
Vx = 43.46 * cos 30° = 37.64 m/s
Vy = 43.46 sin 30° = 21.73 m/s
Distance travelled across the vertical plane,
L = Lo + Vy *t + ½gt²
0 = 86 + 21.73t - 4.9t²
4.9t² - 21.73t - 86 = 0
Solving for t in the quadratic equation,
t = 6.96 or -10.04
Using the positive root since time can't be negative, t = 6.96 approximately 7.0s
Answer:
Here we do not have the vector, but I will try to give a kinda general solution to this type of problem.
If the vector is written as (a, b, c) we have that the force in the x-axis is of a Newtons, in the y-axis is of b Newtons, and in the z-axis is of c Newtons.
Then, we can calculate the total magnitude of this force as:
F = √( a^2 + b^2 + c^2)
wich gives us the total magnitude of the force, but not a direction or anything like that, this is just a scalar.
Answer:
0.739
Explanation:
If we treat the four tire as single body then
W ( weight of the tyre ) = mass × acceleration due to gravity (g)
the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r
where v is speed 25.6 m/s and r is the radius of the circle
centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²
net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²
coefficient of static friction between the tires and the road = frictional force / force of normal
frictional force = m × net acceleration / m×g
where force of normal = weight of the body in opposite direction
coefficient of static friction = (7.2567 × m) / (9.81 × m)
coefficient of static friction = 0.739
