Answer:
9.6 m
Explanation:
This is a case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .
a = 3.6 t + 5.6
d²x / dt² = 3.6 t + 5.6
Integrating on both sides
dx /dt = 3.6 t² / 2 + 5.6 t + c
where c is a constant.
dx /dt = 1.8 t² + 5.6 t + c
when t = 0 , velocity dx /dt is zero
Putting these values in the equation above
0 = 0 +0 + c
c = 0
dx /dt = 1.8 t² + 5.6 t
Again integrating on both sides
x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁
x = 0.6 t³ + 2.8 t² + c₁
when t =0, x = 0
c₁ = 0
x = 0.6 t³ + 2.8 t²
when t = 1.6
x = .6 x 1.6³ + 2.8 x 1.6²
= 2.4576 + 7.168
= 9.6256
9.6 m
By Newton's second law, we have
So, in order to give a 0.15kg body an acceleration of 40m/s^2, you need a force of
Answer:
t = 0.37 seconds
Explanation:
t = (1/4)T
Maximum acceleration is;
a_max = Aω²
In simple harmonic motion, we know that v_max = Aω
Thus, a_max = v_max•ω
ω = a_max/v_max
We know that Period is given by;
T = 2π/ω
From initially, t = (1/4)T so, T = 4t
Thus, 4t = 2π/(a_max/v_max)
t = (2π/4)(v_max/a_max)
We are given;
Maximum velocity;v_max = 1.47 m/s
Max acceleration;a_max =6.24 m/s²
Thus,
t = (2π/4)(1.47/6.24)
t = 0.37 seconds
The equation that relates the voltages and the number of turns in a transformer is
where
is the voltage in the secondary coil,
is the voltage in the primary coil, and
and
are the number of turns on the secondary and primary coils.
Using the numbers, we find the ratio between the number of turns:
