Answer:
2s
Explanation:
It takes the car 1 second to decelerate 12m and it decelerated 24m (as 36 - 12 = 24)
Answer:
Speed of the car, v = 8.90 m/s
Explanation:
It is given that,
The centripetal acceleration experienced by the car,
Diameter of the circle, d = 27 m
Radius of the circle, r = 13.5 m
The centripetal acceleration is experienced by an object if it moves in a circular path. The formula for the centripetal acceleration is given by :
v = 8.90 m/s
So, the maximum speed of the car in this roundabout is 8.9 m/s. Hence, this is the required solution.
Answer:
y = 428.67 m and x all = 1513.68 m
Explanation:
This problem of kinematics can be divided into two parts: a first part when the rockets work a second as a parabolic launch.
Let's do the first part, let's calculate the speed just when the engines turn off
vf = v₀ + at
vf = 78 + at
vf = 78 +12 3
vf = 114 m / s
This is the speed with which the second part begins vo = 114 m / s with an Angle of 38º
Also at this time a distance is displaced, we calculate the distance traveled (in the direction of the acceleration)
d = v₀ t + ½ a t²
d = 78 3 + ½ 12 3²
d = 288 m
Let's use trigonometry to find the components
x₀ = d cos 38 = 288 cos 38
y₀ = d sin38 = 288 sin38
x₀ = 226.95 m
y₀ = 177.31 m
Second part
Let's calculate the maximum height, at this point its vertical speed is zero (vfy = 0)
Let's decompose the initial velocity using trigonometry
vₓ = v₀ cos 38
= v₀ sin38
vₓ = 114 cos 38
= 114 sin38
vₓ = 89.83 m / s
= 70.19 m / s
² = ² - 2g (y -y₀)
0 = ² -2g (y -yo)
y-y₀ = ² / 2g
y-y₀ = 70.19²/2 9.8
y = 251.36 + y₀
y = 251.36 + 177.31
y = 428.67 m
This is the maximum height from the point where the movement began, that is, the ground.
Now let's calculate the range
R = vo² sin 2θ / g
R = 114² sin 2 38 /9.8
R = 1286.73 m
This is the scope of the parabolic movement, we must add the horizontal distance traveled in the first part
x all = R + xo
x all = 1286.73 + 226.95
x all = 1513.68 m
Answer:
solved
Explanation:
a) F_net = (F2 - F3)i - F1 j
b) |Fnet| = sqrt( (F2 - F3)^2 + F1^2)
= sqrt( (9- 5)^2 + 1^2)
= 4.123 N
c) θ = tan^-1( (Fnet_y/Fnet_x)
= tan^-1( -1/(9-5) )
= -14.036°