Answer:
9m^3
Explanation:
Given data
volume v1= 3m^3
volume v2= ???
Temperature T1= 20.0°C.
Temperature T2= 60.0°C.
Applying the relation for temperature and volume
V1/T1= V2/T2
substitute
3/20= V2/60
3*60= V2*20
180= 20*V2
180/20= V2
V2= 9m^3
Hence the final volume is 9m^3
The object increases in speed
I KNOW that you must have seen this formula before. It gives
the distance covered in a certain time of accelerated motion:
D = 1/2 A T²
Distance covered= (1/2) (Acceleration) (Time squared) .
The question gives us the acceleration and the time.
I've got a weird idea: Let's plug them into the formula. OK ?
Distance = (1/2) (10 m/s²) (10 sec)²
= (5 m/s²) (100 sec²)
= 500 meters .
Deep lake !