Answer:
v = 6.45 10⁻³ m / s
Explanation:
Electric force is
F = q E
Where q is the charge and E is the electric field
Let's use Newton's second law to find acceleration
F- W = m a
a = F / m - g
a = q / m E g
Let's calculate
a = -1.6 10⁻¹⁹ / 9.1 10⁻³¹ (-1.30 10⁻¹⁰) - 9.8
a = 0.228 10² -9.8
a= 13.0 m / s²
Now we can use kinematics, knowing that the resting parts electrons
v² = v₀² + 2 a y
v =√ (0 + 2 13.0 1.6 10⁻⁶)
v = 6.45 10⁻³ m / s
As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards
So here we know that
now from the above equation
so both of the charges will apply 0.288 N force on q3 charge along the line joining them
now the net force due to vector sum is given by
here we know that angle is
now we have
so net force on q3 is 0.46 N vertically upwards along +Y axis
<h2>Answer</h2>
option D)
2.4 seconds
<h2>Explanation</h2>
Given in the question,
mass of car = 1200kg
speed of car = 19m/s
Force due to direction of travel
F = ma
= 12000(a)
Force to due frictional force in reverse direction
-F = mg(friction coefficient)
= -12000(9.81)(0.8)
<h2>
-mg(friction coefficient) = ma </h2>
(cancelling mass from both side of equation)
g(0.8) = a
(9.81)(0.8) = a
a = 7.848 m/s²
<h2>Use Newton Law of motion</h2><h3>vf - vo = a • t</h3>
where vf = final velocity
vo = initial velocity
a = acceleration
t = time
0 - 19 = 7.8(t)
t = 19/7.8
= 2.436 s
≈ 2.4s
