A 1 490-kg automobile has a wheel base (the distance between the axles) of 3.10 m. The automobile's center of mass is on the centerline at a point 1.10 m behind the front axle. Find the force exerted by the ground on each wheel.

Force on each front wheel = 4711 N

Force on each back wheel = 2591 N

Explanation:

Weight of car = 1490 Kg

Distance of COG from front axle= 1.1 m

Distance of COG from back axle = 3.1 - 1.1 = 2 m

Vertical forces are balanced ,

So, R1 + R2 = mg

R1 + R2 = 1490× 9.8

Thus; R1 + R2 = 14602 - - - eq1

Where

R1 is force on front wheel

R2 is force on back wheel

Now, we know that;

Angular momentum about centre of gravity is zero.

Therefore,

R1 × 1.1 = R2 × 2

R1 = (2/1.1)R2

Put this into equation1, to get;

(2/1.1)R2 + R2 = 14602

R2(2.818) = 14602

R2 = 14602/2.818

R2 = 5182 N and R1 = (2/1.1)5182 = 9422 N

Hence, force on each back wheel = R2/2 = 5182/2 = 2591 N

Force on each front wheel = R1/2 = 9422/2 = 4711 N

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j