Answer:
v=32.9m/s
Explanation:
The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation 
This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).
We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:
Answer:
1.73 m/s²
3.0 cm
Explanation:
Draw a free body diagram of the yo-yo. There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.
Sum of forces in the y direction:
∑F = ma
T cos 10° − mg = 0
T cos 10° = mg
T = mg / cos 10°
Sum of forces in the x direction:
∑F = ma
T sin 10° = ma
mg tan 10° = ma
g tan 10° = a
a = 1.73 m/s²
Draw a free body diagram of the sphere. There are two forces: weight force mg pulling down, and air resistance D pushing up. At terminal velocity, the acceleration is 0.
Sum of forces in the y direction:
∑F = ma
D − mg = 0
D = mg
½ ρₐ v² C A = ρᵢ V g
½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g
3 ρₐ v² C = 8 ρᵢ r g
r = 3 ρₐ v² C / (8 ρᵢ g)
r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))
r = 0.030 m
r = 3.0 cm
The circuit is no longer closed.
Answer: vf = 51 m/s
d = 112 m
Explanation: Solution attached:
To find vf we use acceleration equation:
a = vf - vi / t
Derive to find vf
vf = at + vi
Substitute the values
vf = 3.5 m/s² ( 8.0 s) + 23 m/s
= 51 m/s
To solve for distance we use
d = (∆v)² / 2a
= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)
= (28 m/s)² / 7 m/s²
= 784 m/s / 7 m/s²
= 112 m
Answer:
The car stops after 32.58 m.
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 20 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -6 m/s²
Time taken by the car to stop
Total Time taken by the car to stop is 0.5+3.33 = 3.83 s
The car stops after 32.58 m.
Distance between car and obstacle is 50-32.58 = 17.42 m
