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:
Explanation:
Initial angular velocity ω₀ = 151 x 2π / 60
= 15.8 rad /s
final velocity = 0
Angular deceleration α = 2.23 rad / s
ω² = ω₀² - 2 α θ
0 = 15.8² - 2 x 2.23 θ
= 55.99 rad
one revolution = 2π radian
55.99 radian = 55.99 / 2 π no of terns
= 9 approx .
Mechanical and electrical
Answer:
(a) 7.72×10⁵ J
(b) 4000 J
(c) 1.82×10⁻¹⁶ J
Explanation:
Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,
Ek = 1/2mv²................... Equation 1
Where Ek = Kinetic energy, m = mass, v = velocity
(a)
For a moving automobile,
Ek = 1/2mv².
Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s
Substitute into equation 1
Ek = 1/2(2.0×10³)(27.78²)
Ek = 7.72×10⁵ J
(b)
For a sprinting runner,
Given: m = 80 kg, v = 10 m/s
Substitute into equation 1 above,
Ek = 1/2(80)(10²)
Ek = 40(100)
Ek = 4000 J
(c)
For a moving electron,
Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s
Substitute into equation 1 above,
Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²
Ek = 1.82×10⁻¹⁶ J
