Answer
Assuming
east is the positive x direction
north is the positive y direction
initial velocity , u = 19 j m/s
a)
acceleration , a = 1.6 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 + 5.6× 1.6
v = 28 j m/s
the velocity of the car after 5.6 s is 28 m/s north
b)
acceleration , a = -1.5 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 - 5.6 ×1.5
v = 10.6 j m/s
the velocity of the car after 5.6 s is 10.6 m/s north
Answer:
t = 22.2 s
Explanation:
angular distance covered in the 36.0 s is
θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians
720/2 = 360 radians
α = Δω/t = (30 - 10)/36 = 5/9 rad/s²
θ = ω₀t + ½αt²
360 = 10.0t + ½(5/9)t²
0 = (5/18)t² + 10.0t - 360
0 = t² + 36t - 1296
t = (-36 ±√(36² - 4(1)(-1296))) / 2
t = (-36 ±√(6480)) / 2
t = -18 ±√1620
we ignore the negative time result as it occurs before we care.
t = -18 + √1620 = 22.249223... s
The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2) where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.
Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span>
The car's speed is 240km/4hr= 60km/hr.
There's not enough information given in the question to determine its velocity.