Answer:
a = 1.055 x 10¹⁷ m/s²
Explanation:
First, we will find the force on electron:
where,
F = Force = ?
E = Electric Field = 6 x 10⁵ N/C
q = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
F = 9.6 x 10⁻¹⁴ N
Now, we will calculate the acceleration using Newton's Second Law:
where,
a = acceleration = ?
m = mass of electron = 9.1 x 10⁻³¹ kg
therefore,
<u>a = 1.055 x 10¹⁷ m/s²</u>
Answer:
a) 10.51 J
b) 3.48 m/s
Explanation:
Given data :
mass of train ( M ) = 2.2 kg
Given initial velocity ( u ) = 1.6 m/s
<u>a) calculating work done by the force over the journey of the train</u>
F = mx + b ------ ( 1 )
m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m
x = distance travelled on the x axis by the train = 7.5 m
F = force experienced by the train = 2.8 N
x = 0
∴ b = 2.8
hence equation 1 can be written as
F = ( -0.373) x + 2.8 ----- ( 2 )
hence to determine the work done by the force
W = Note: the limits are actually 7.5 and 0
∴ W ( work done ) = -10.49 + 21 = 10.51 J
<u>b) calculate the speed of the train at the end of its journey</u>
we will apply the work energy theorem
W = 1/2 m*v^2 - 1/2 m*u^2
∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )
V^2 = 12.11
hence V = 3.48 m/s