Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.300 m. The engine of the car applies a torque of 222 N·m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

740 N

Explanation:

We are given that

Radius,r=0.3 m

Torque,

We have to find the magnitude of the static frictional force.

According to question

Torque by engine=Torque by static friction

Hence, the magnitude of static frictional force=740 N

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

answer;

d. are made mostly of liquid

explanation;

-the general   composition and structure planets includes; they have a thick gas and liquid layer covering a small,solid core. mostly the gases are hydrogen and helium atmosphere and layers of liquid hydrogen. for instance neptune and uranus are examples of the outer planets, they have a core covered by ice and ammonia.

-neptune's atmosphere is hydrogen and helium with a bit of methane so there is no oxygen to breath and the interior is made up of frozen water and ammonia with a rock and iron core. thus, there is no liquid water or solid surface to survive.