Answer:
0.1 s
Explanation:
The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log
So F - f = ma
F - μmg = ma
F/m - μg = a
So, substituting the values of the variables into the equation, we have
a = F/m - μg
a = 2850 N/300 kg - 0.45 × 9.8 m/s²
a = 9.5 m/s² - 4.41 m/s²
a = 5.09 m/s²
Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.
So, making t subject of the formula, we have
t = (v - u)/a
substituting the values of the variables into the equation, we have
t = (v - u)/a
t = (0.5 m/s - 0 m/s)/5.09 m/s²
t = 0.5 m/s ÷ 5.09 m/s²
t = 0.098 s
t ≅ 0.1 s
<span>Assuming the car is travelling in the same direction for the entire hour, the acceleration is zero.</span>
<span>10 inches
You are at risk of serious injury if you sit less than 10 inches away from the steering wheel, because of the speed and force the airbag deploys at. This is also part of the reason why driving instructors now instruct you to hold the steering wheel from the lower parts, rather than the top, which can cause your thumbs to break if the air bag deploys.</span>
Answer:
l= 3.002 cm
Explanation:
Given that
n= 70 turns
B= 1.2 T
θ= 15°
I= 1.5 A
τ = 0.0294 N⋅m
Lets take length of sides is l.
We know that
τ = n I A B sin θ
Area of square ,A= l²
Now by putting the value
τ = n I A B sin θ
0.0294 = 70 x 1.5 x l² x 1.2 x sin 15°
l² = 0.000901 m²
l² = 9.01 cm²
l= 3.002 cm
