Answer:
Both the frequency f and velocity v will increase.
When the radius reduces, the circumference of the circular path becomes smaller which means that more number of revolutions can be made per unit time as long as the force is kept constant; this is an increase in frequency.
Explanation:
The centripetal force acting on a mass in circular motion is given by equation (1);
where m is the mass of the object and r is radius of the circle. From equation one we see that the centripetal force is directly proportional to the square of the velocity and inversely proportional to the radius of the circular path.
However, according to the problem, the force is constant while the radius and the velocity changes. Therefore we can write the following equation;
Also recall that m is constant so it cancels out from both sides of equation (2). Therefore from equation we can write the following;
By observing equation (2) carefully, the ratio 
will with the square root increase 
since 
is lesser than 
.
Hence by implication, the value of 
will be greater than .
As the radius changes from to , the velocity also changes from to .
Answer:
μk = (Vf - Vc)/(T×g)
Explanation:
Given
Vi = initial velocity of the car
Vf = final velocity of the car
T = Time of application of brakes
g = acceleration due to gravity (known constant)
Let the mass of the car be Mc
Assuming only kinetic frictional force acts on the car as the driver applies the brakes,
The n from Newtown's second law of motion.
Fk = Mc×a
Fk = μk×Mc×g
a = (Vf - Vc)/T
Equating both preceding equation.
μk×Mc×g = Mc × (Vf - Vc)/T
Mc cancels out.
μk = (Vf - Vc)/(T×g)
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Answer
is: V<span>an't
Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to
solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per
kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).
i = 1,81.
