The coefficient of static friction is 0.222
Explanation:
In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:
where the term on the left is the frictional force, while the term on the right is the centripetal force, and where
is the coefficient of static friction
m is the mass of the car
g is the acceleration of gravity
v is the speed of the car
r is the radius of the track
In this problem, we have:
r = 564 m
v = 35 m/s
And re-arranging the equation for , we can find the coefficient of static friction:
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Answer:
4.7 m³
Explanation:
We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2
* Givens :
P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K
( We must always convert the temperature unit to Kelvin "K")
* What we want to find :
V2 = ?
* Solution :
101 × 2 / 300.15 = 40 × V2 / 283.15
V2 × 40 / 283.15 ≈ 0.67
V2 = 0.67 × 283.15 / 40
V2 ≈ 4.7 m³
Answer:
The object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g.
No so sure
Explanation:
Hope it helps
