Answer:
acceleration of the car is 3 m\s^2
Explanation:
from rest means the initial velocity (vi) is zero
time = 5s
final velocity (vf) = 15m\s
a = vf - vi \ t
a = (15-0) \ 5
a= 3 m\s^2
which means that the car is speeding up 3 meters every second
Gravity is counteracted by centripetal force, due to acceleration, which is the force that pushes you into your seat.
To solve the problem it is necessary to apply conservation of the moment and conservation of energy.
By conservation of the moment we know that
Where
M=Heavier mass
V = Velocity of heavier mass
m = lighter mass
v = velocity of lighter mass
That equation in function of the velocity of heavier mass is
Also we have that 
On the other hand we have from law of conservation of energy that
Where,
W_f = Work made by friction
KE = Kinetic Force
Applying this equation in heavier object.
Here we can apply the law of conservation of energy for light mass, then
Replacing the value of 
Deleting constants,
Answer:
scatter plots show the relationship between the independent and dependent variables
Explanation:
A scatter plot is a graph which shows two variables plotted along two axes (usually the x and y axes). Scatter plots are useful in establishing any form of correlation between the dependent and independent variables in any study.
Correlation simply means the degree of relationship between variables, that is, how much does one variable affect the other? When scatter plots are almost a straight line graph, there is a high correlation between the variables. When the points in a scatter plot are isolated, there is little (sometimes zero) correlation between the variables.
Answer:
I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.
Explanation:
In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:
R = V₀² Sin 2θ/g
where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.
The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:
V₀ = √(V₀ₓ² + V₀y²)
where,
V₀ₓ = Horizontal Velocity
V₀y = Vertical Velocity
Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.
<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>
