Answer:
The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"
answers:
a.
b. 
Explanation:
First let state the mathematical expression for the tangential acceleration and the radial acceleration.
a. tangential acceleration is express as

since the distance is expressed as

the derivative is the velocity, hence

hence when we take the drivative of the velocity we arrive at
b. the expression for the radial acceleration is expressed as

-- First, we have to decide how to handle the two resistors.
The effective resistance of resistors in series is the sum
of their individual resistances. That is, they act like a single
resistor, whose resistance is the sum of all of them.
So in this question, the 4.0 ohms and the 7.5 ohms act like a
single resistor of 11.5 ohms.
-- The current in the circuit is
(the supply voltage) / (the total resistance)
= (9.0 volts) / (11.5 ohms)
= 0.783... Ampere (rounded)
This behavior helps Betty in <u>intellectual </u>development.
Answer:
Final velocity = 7.677 m/s
KE before crash = 202300 J
KE after crash = 182,702.62 J
Explanation:
We are given;
m1 = 1400 kg
m2 = 4700 kg
u1 = 17 m/s
u2 = 0 m/s
Using formula for inelastic collision, we have;
m1•u1 + m2•u2 = (m1 + m2)v
Where v is final velocity after collision.
Plugging in the relevant values;
(1400 × 17) + (4700 × 0) = (1400 + 1700)v
23800 = 3100v
v = 23800/3100
v = 7.677 m/s
Kinetic energy before crash = ½ × 1400 × 17² = 202300 J
Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J