Answer:
F= 4788 N
Explanation:
Because the car moves with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
d=36.9 m
v₀=14.0 m/s m/s
vf= 0
Calculating of the acceleration of the car
We replace dta in the formula (1)
vf²=v₀²+2*a*d
(0)²=(14)²+2*a*(36.9)
-(14)²= (73.8) *a
a= - (196) / (73.8)
a= - 2.66 m/s²
Newton's second law of the car in direction horizontal (x):
∑Fx = m*ax Formula (2)
∑F : algebraic sum of the forces in direction x-axis (N)
m : mass (kg)
a : acceleration (m/s²)
Data
m=1800 Fkg
a= - 2.66 m/s²
Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :
We replace data in the formula (2)
-F= (1800 kg) * ( -2.66 m/s²
)
F= 4788 N
Answer:
The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, w = mg.
<span>I of disk = 1/2MR² = (0.5)(230)(4.0)² = 1840 kgm²
I of disk-woman system = I = 1/2MR² + md² = (50)(4.0)² = 1840 + 800 = 2640 kgm²
L = Iw = (2640)(0.80)(2π) = 13,270 ≈ = 13,000 kgm²/s ANS
idk I'm just trying</span>