Answer:
9. A 1500kg car traveling +6m/s with a 2000kg truck at rest. The vehicles collide, but do not stick together. The car has a velocity -3m/s after the collision. What is the velocity of the truck? a. What type of collision occurred above?
Answer:
d = V/E
Explanation:
From the definition, we can say that the electric field strength between the plates of a parallel plate capacitor is
E = v/d
where
E = electric field strength
V = potential difference
d = distance between the plates
On rearranging the equation and making d subject of the formula, we have
d = V/E
From the question, we're given that
V = 112 V
E = 1.12 kV/cm converting to V/m, we have 110000 V/cm
d = 112 / 110000
d = 0.00102 m
d = 1.02*10^-3 m
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :

1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
<em>The </em><em>nucleus</em><em> has most of the atomic mass in an atom. The </em><em>nucleus</em><em> is made up of protons and neutrons.</em>
<em />
You have to divide the pressure exerted by the air into two partial pressures: of the dry air and of the water vapor. Combining these two values gives you the parameter.