A) How many meters are there in 7.50 light-year? 7.10*10^16 2) An unmarked polic

Question

A) How many meters are there in 7.50 light-year? 7.10*10^16

2)An unmarked police car, traveling a constant 91km/h , is passed by a speeder traveling 158km/h . Precisely 1.00s after the speeder passes, the policeman steps on the accelerator.

A) If the police car's acceleration is 1.90m/s^2 , how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at constant speed)? Express your answer using three significant figures.

3) An average family of four uses roughly 1200 liters (about 300 gallons) of water per day. (One liter = 1000 cm^3)

A) How much depth would a lake lose per year if it uniformly covered an area of 50 square kilometers and supplied a local town with a population of 70000 people? Consider only population uses, and neglect evaporation and so on.

Explanation / Answer

(B) 2 light years = 1.89*10^16 m

1 AU = 1.5*10^11 m

So, no. of AUs = 1.89*10^16 / 1.5*10^11 = 126000 AUs.

.

.

.

(2) 91 km/hr = 25.278 m/s... 158 km/hr = 43.889 m/s....

So, relative velocity of speeder w.r.t police = 43.889-25.278 = 18.611 m/s....

After 1 sec, distance between them would be = 18.611 m....

Let they meet after time 't'.

Distance travelled by police = distance traveled by speeder + 18.611

1/2 * 1.9*t^2 = 18.611*t + 18.611

t = 20.544 seconds.

So, he meets him 20.544 seconds after he starts accelerating.

.

.

.

(3) Amount of water used per day by 70000 people = 1200/4 * 70000 = 21*10^6 liters = 21000 m^3

So, volume used per year = 7665000 m^3

Area of lake = 50 km^2 = 5*10^7 m^2

So, depth decreased = 7665000/5*10^7 = 0.1533 m = 15.33 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.