Basically, when someone is resting in an accelerated vehicle without restraint from a seatbelt, the force of stopping the vehicle will be when inertia occurs, and that force of the vehicle coming to a stop will affect the passenger (without a seatbelt/restraint from another force or object) greatly by throwing them.
For example;
If I were to be riding in a vehicle (without a seatbelt) that's accelerating at 40 m/s^2 and it suddenly gets slammed on the breaks, I will be thrown forward from inside the vehicle.
I hope this helps!
Answer:
x = 25 / μ [ ft]
Explanation:
To solve this exercise we can use Newton's second law.
Let's set a reference system where the x axis is parallel to the road
Y axis
N_B + N_A - W_van - W_load = 0
N_B + N_A = W_van + W_load
X axis
fr = ma
a = fr / m
the total mass is
m = (W_van + W_load) / g
the friction force has the expression
fr = μ N_{total}
fr = μy (W_van + W_load)
we substitute
a = μ (W_van + W_load) 
a = μ g
taking the acceleration let's use the kinematic relations where the final velocity is zero
v² = v₀² - 2 a x
0 = v₀² -2a x
x = 
x = 
x = 
x = 25 / μ [ ft]
Answer:
Heat
Explanation:
Because chemical energy is stored, it is a form of potential energy. When a chemical reaction takes place, the stored chemical energy is released. Heat is often produced as a by-product of a chemical reaction – this is called an exothermic reaction.
Hope this helped.
Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
