Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s
Answer:
The rate of the boat in still water is 44 mph and the rate of the current is 4 mph
Explanation:
x = the rate of the boat in still water
y = the rate of the current.
Distance travelled = 120 mi
Time taken upstream = 3 hr
Time taken downstream = 2.5 hr
Speed = Distance / Time
Speed upstream

Speed downstream

Adding both the equations


The rate of the boat in still water is <u>44 mph</u> and the rate of the current is <u>4 mph</u>
Answer:
a) Beth will reach before Alan
b)Beth has to wait 20 min for Alan to arrive
Explanation:
let 'd' be distance b/w Los Angeles and San Francisco i.e 400 mi
considering ,
Alan's speed
=50mph
Beth's speed
=60mph
->For Alan:
The time required
= d/
= 400/50 => 8h
-> For beth:
The time required
=> 6h 40m
Alan will reach at 8:00 a.m +8h = 4:00p.m.
Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.
a) Beth will reach before Alan
b)Beth has to wait 20 min for Alan to arrive
Answer:

Explanation:
Given that,
The car traveled a total of 1,200 meters during this test.
We need to find the average speed of the car. The average speed of the car is given by total distance covered divided by the time taken. So,

But putting the value of t we can find the average speed of the car.