The coefficient of friction must be 0.196
Explanation:
For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:
where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:
is the coefficient of friction between the tires and the road
m is the mass of the car
is the acceleration of gravity
v is the speed of the car
r is the radius of the curve
In this problem,
r = 750 m is the radius
is the speed
And solving for
, we find the coefficient of friction required to keep the car in circular motion:

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Answer:
Balances and Scales
A balance compares an object with a known mass to the object in question. One example of a balance is the triple beam balance. The standard unit of measure for mass is based on the metric system and is typically denoted as kilograms or grams.
Answer:
Correct answer: Kinetic energy of rotation, kinetic energy of translation and potential energy
Explanation:
Before it began to rotate it possessed potential energy. After being released from the rest it starts to roll on the roof and get kinetic energy of rotation, kinetic energy of translation along with potential energy.
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Answer:
a) 0.477 W/m²
b) 13.407 N/C
c) 18.96 N/C
Explanation:
P = Power = 150 W
r = Distance = 5 m
ε₀ = Permittivity of space = 8.854×10⁻¹² F/m
a) Average intensity

∴ Average intensity is 0.477 W/m²
b) Rms value

∴ Rms value of the electric field is 13.407 N/C
c) Peak value

∴ Peak value of the electric field is 18.96 N/C
Quartz, gold and calcite are examples of minerals, coal is a fossil fuel