S= 343m/s
F=256Hz
WL= 343ms/256-1
WL=V/F
= 1.339844m
<h3><u>Answer;</u></h3>
volume = 6.3 × 10^-2 L
<h3><u>Explanation</u>;</h3>
Volume = mass/density
Mass = 0.0565 Kg,
Density = 900 kg/m³
= 0.0565 kg/ 900 kg /m³
= 6.3 × 10^-5 M³
but; 1000 L = 1 m³
Hence, <u>volume = 6.3 × 10^-2 L</u>
Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time, 
(a) Acceleration, 
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :


s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration, 


s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.
Answer:
a = 1 m/s² and
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
Ty - W = 0
Ty = w
X axis
Tx = m a
With trigonometry we find the components of tension
Sin θ = Ty / T
Ty = T sin θ
Cos θ = Tx / T
Tx = T cos θ
We calculate the acceleration with kinematics
Vf = Vo + a t
a = (Vf -Vo) / t
a = (20 -10) / 10
a = 1 m/s²
We substitute in Newton's equations
T Sin θ = mg
T cos θ = ma
We divide the two equations
Tan θ = g / a
θ = tan⁻¹ (g / a)
θ = tan⁻¹ (9.8 / 1)
θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle