Answer:
a=0 v = v₀ + a t
a=0 line is horizontal
Explanation:
1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration
2, speed and relationship of a car is given by
v = v₀ + a t
where vo is the initial velocity, a is the acceleration and tel time
in this case I will calcograph velocity vs. time the constant acceleration is a straight line.
In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope
In the given problem, we say various information's that are going to help us reach the ultimate answer to the question. Let us first write the information's that have been presented in front of us.
Mass of the car = 2000 kg
Velocity of the car = 25 m/s^2
Radius of the circle = 80 m
Now we already know the equation for calculating the centripetal force and that is
Centripetal Force = [mass * (velocity)^2]/Radius
= [2000 * (25)^2]/80
= (2000 * 625)/80
= 1250000/80
= 15625
So the centripetal force on the car is 15625 Newtons
According to periodic trends in the periodic table, the atomic radius decreases from left to right.
In period three, the element with the smallest atomic radius would be the element in the rightmost area. Protons increase as it goes to the right, which would mean they pull in electrons closer which decreases the size.
So in period 3, the element with the smallest atomic radius is Argon (Ar).
Answer:
<h3>
2.3125m/s²</h3>
Explanation:
Using the equation of motion v² = u²+2aS
v is the final velocity = 120km/hr
120km/hr = 120 * 1000/1 * 3600 = 33.3m/s
u is the initial velocity = 0m/s
a is the acceleration
S is the distance covered = 240m
On substituting the given parameters
33.3² = 0²+2a(240)
33.3² = 480a
1110 = 480a
a = 1110/480
a = 2.3125m/s²
Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²
