Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
Answer: The net force acting on the car 1,299.3 N.
Explanation:
Mass of the car = 710 kg
Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s 
Final velocity of the car,v = 120 km/h = 33.33 m/s
time taken b y car = 12.6 sec
v-u=at
The net force acting on the car 1,299.3 N.
Answer:
a) = 258352.5J
b) = 23.63 m/s
c) = 1.8m
Explanation:
Data;
Mass = 925kg
Distance (s) = 28.5m
Force constant (k) = 8.0*10⁴ N/m
g = 9.8 m/s²
a) = work = force * distance
But force = mass * acceleration
Force = 925 * 9.8 = 9065N
Work = F * s = 9065 * 28.5 = 258352.5J
b) acceleration (a) = (v² - u²) / 2s
a = v² / 2s
v² = a * 2s
v² = 9.8 * (2 * 28.5)
v² = 9.8 * 57
v² = 558.6
v = √(558.6)
V = 23.63 m/s
C). The work stops when the work done to raise the spring equals the work done to stop it by the spring
W = ½kx²
258352.5 = ½ * 8.0*10⁴ * x²
(2 * 258352.5) = 8.0*10⁴x²
516705 = 8.0*10⁴x²
X² = 516705 / 8.0*10⁴
X² = 6.46
X = √(6.46)
X = 2.54m
The compression was about 2.54m
Answer:
Torque = 35.60 N.m (rounded off to 3 significant figures.
Explanation:
Given details:
The mass of the rock on the left, ms = 2.25 kg
The total mass of the rocks, mp = 10.1 kg
The distance from the fulcrum to the center of the pile of rocks, rp = 0.360 m
(a) The torque produced by the pile of rock, T = F*rp = m*g*rp
Torque = 9.8*0.360*10.1 = 35.6328
Torque = 35.60 N.m (rounded off to 3 significant figures).
