Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:

where F is the magnitude of the force, m is the mass of the object and a its acceleration.
In this problem, the object is the insect, with mass

. The acceleration of the insect is

, therefore we can calculate the force exerted by the car on the insect:

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
Cuando la presión aumenta en el agua, disminuye el punto de fusión del hielo. Ósea esta a temperatura muy alta.
Espero que esto te ayude y no me haya confundido