I say around 40% - 60%
https://www.dmv.ca.gov/portal/dmv/detail/teenweb/more_btn6/traffic/traffic
http://www.teendriversource.org/stats/support_teens/detail/57
http://www.rmiia.org/auto/teens/Teen_Driving_Statistics.asp
(I just corrected the question. Sorry if it is still incorrect.)
a.
The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:
where F is the magnitude of the force, theta is the angle between them and d is the distance.
The problen gives the following data:
The magnitude of the force 750 N.
The angle between the force and the displacement which is 25°
The distance, 26 m.
Plugging this in the formula we have:
Therefore the work done is 17673 J.
b)
The power is given by:
the problem states that the time it takes is 6 s. Then:
Therefore the power is 2945.5 W
Answer:
The most common oxidation numbers for a given element
Answer:
(a) m = 33.3 kg
(b) d = 150 m
(c) vf = 30 m/s
Explanation:
Newton's second law to the block:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Data
F= 100 N
a= 3.0 m/s²
(a) Calculating of the mass of the block:
We replace dta in the formula (1)
F = m*a
100 = m*3
m = 100 / 3
m = 33.3 kg
Kinematic analysis
Because the block moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+a*t Formula (3)
Where:
d:displacement in meters (m)
t : time interval in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
a= 3.0 m/s²
v₀= 0
t = 10 s
(b) Distance the block will travel if the force is applied for 10 s
We replace dta in the formula (2):
d= v₀t+ (1/2)*a*t²
d = 0+ (1/2)*(3)*(10)²
d =150 m
(c) Calculate the speed of the block after the force has been applied for 10 s
We replace dta in the formula (3):
vf= v₀+a*t
vf= 0+(3*(10)
vf= 30 m/s
