Horizontal forehead wrinkles are caused by contraction of the frontalis muscles. The frontalis muscles are two large fanlike muscles that extend from the eyebrow region to the top of the forehead. I think this is the answer
Anthropology
Hope this helps! ^-^
<span>Atoms with the same atomic number but different atomic mass are called:
<span>Isotopes</span>
</span>
The normal force is the supporting force that is exerted on an object that is in contact with another stable object.
Answer: Option C
<u>Explanation:
</u>
Normal force is forward or upward pushing force acting on an object. Mostly the normal force acts as supporting force exerted on the object by the neighbouring stable object with which the object in question is in contact. So normal force falls under the category of contact forces.
Generally, normal force will be acting to support the weight of any object placed on another object. The best examples of normal forces are the weight of the book supported by table or by the pushing force of the wall on the person leaning on the wall.
Answer:
(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m
(b) thermal energy was generated by friction is 1.88 x J
(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N
Explanation:
given information:
m = 750 kg
initial velocity, = 110 km/h = 110 x 1000/3600 = 30.6 m/s
initial height, = 22 m
slope, θ = 2.5°
(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?
according to conservation-energy
EP = EK
mgh =
gh =
h =
= 47.6 m
(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?
thermal energy = mgΔh
= mg (h - )
= 750 x 9.8 x (47.6 - 22)
= 188160 Joule
= 1.88 x J
(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?
f d = mgΔh
f = mgΔh / d,
where h = d sin θ, d = h/sinθ
therefore
f = (mgΔh) / (h/sinθ)
= 1.88 x /(22/sin 2.5°)
= 373 N