Answer:
61.76 N.
Explanation:
Given the mass of the car, m = 1.60 kg.
The speed of the car, v = 12.0 m/s.
The radius of the circle, r = 5 m.
As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.
Let N is the normal force.
So, 

Now substitute the given values, we get


N = 61.76 N.
Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.
44/5 seconds to g<span>o from a complete stop to 44 km/h</span>
Answer:
The proportion of flanges that exceeds 0.99 millimeters is 0.6
Explanation:
Given;
integral range of [0.95, 1.05]
Let X be a variable with uniform distribution over the given range.

1 - 0.95 = 0.05, 1.05 - 1 = 0.05

interval = 10 - 0.5 = 9.5
F(x) = 10x + 9.5
When, X exceeds 0.99 millimeters, then the proportion of flanges will be;
P (X > x) = 1 - F(x)
P( X > 0.99 ) = 1 - 10(0.99) + 9.5
P( X > 0.99 ) = 0.6
Therefore, the proportion of flanges that exceeds 0.99 millimeters is 0.6
Answer:
the voltage across each resistor is one third of the battery voltage
Explanation:
In a series circuit, the current is constant throughout the circuit, so the battery voltage is equal to the sum of the voltage drop in each part of the series circuit.
V = i (R₁ + R₂ + R₃)
in the exercise indicate that all resistance has the same value
R₀ = R₁ = R₂ = R₃
V = i 3 R₀
V
= 3 V₀
V₀ = i R₀
V₀= V / 3
the voltage across each resistor is one third of the battery voltage