Answer:
a) The centripetal acceleration of the car is 0.68 m/s²
b) The force that maintains circular motion is 940.03 N.
c) The minimum coefficient of static friction between the tires and the road is 0.069.
Explanation:
a) The centripetal acceleration of the car can be found using the following equation:

Where:
v: is the velocity of the car = 51.1 km/h
r: is the radius = 2.95x10² m

Hence, the centripetal acceleration of the car is 0.68 m/s².
b) The force that maintains circular motion is the centripetal force:

Where:
m: is the mass of the car
The mass is given by:

Where P is the weight of the car = 13561 N

Now, the centripetal force is:

Then, the force that maintains circular motion is 940.03 N.
c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:



Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.
I hope it helps you!
The speed of the block when the compression is 15 cm is 9.85 m/s.
The given parameters;
- <em>mass of the block, m = 2.4 kg</em>
- <em>height of the block, h = 5 m</em>
- <em>compression of the spring, x = 25 cm = 0.25 m</em>
The spring constant is calculated as follows;

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.
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The answer is Concave lenses
Answer:Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.
Explanation: