Answer:
Explanation:
Before it hits the ground:
The initial potential energy = the final potential energy + the kinetic energy
mgH = mgh + 1/2 mv²
gH = gh + 1/2 v²
v = √(2g (H - h))
v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))
v ≈ 2.0 m/s
When it hits the ground:
Initial potential energy = final kinetic energy
mgH = 1/2 mv²
v = √(2gH)
v = √(2 * 9.81 m/s² * 0.42 m)
v ≈ 2.9 m/s
Using a kinematic equation to check our answer:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)
v ≈ 2.9 m/s
Answer:
<u><em>First Reaction:</em></u>
=> 
<u><em>Second Reaction:</em></u>
=> 
<u><em>Combined Reaction:</em></u>
=> 
Answer:
0.08 N/C
Explanation:
Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,
E = Kq/r².............................. Equation 1
Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.
making q the subject of the equation,
q = Er²/k............................... Equation 2
Given: E = 2 N/C, r = 4 m,
Substitute into equation 2
q = 2(4)²/k
q = 32/k C.
When r is increased to 20 m,
E = k(32/k)/20²
E = 32/400
E = 0.08 N/C.
Hence the electric Field = 0.08 N/C
You can’t just do this tm u have to do it now
It is C) because that truck has more mass
