Let volume of empty boat be = 100% = 1V
and mass of boat be M
In water 10%, 0.1V of the volume is submerged.
Mass, m of 1200kg increases the submerging from 10%, 0.1V to 70%, 0.7V
M leads to 0.1V boat submerging
boat submerging.
M + 1200kg leads to 0.7V boat submerging.
This is 60%, 0.6 V increase
By comparison
(M+1200kg) * 0.1V = 0.7V * M
0.1M + 120kg = 0.7M
120kg = 0.7M - 0.1M
120kg = 0.6M
M = (120/0.6)kg
M = 200kg.
The mass of the boat is 200kg.
Answer:
ρ = 830.32 kg/m³
Explanation:
Given that
Oil head = 12.2 m
h= 12.2 m
Pressure P = 1.013 x 10⁵ Pa
Lets take density of the liquid =ρ
The pressure due to liquid P given as
P = ρ g h
Now by putting the all values in the above equation
1.013 x 10⁵ Pa = ρ x 10 x 12.2 ( take g =10 m/s²)
ρ = 830.32 kg/m³
Therefore the density of oil is 830.32 kg/m³
Answer: A.AB
Explanation:
This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.
As we can see in the attached image, the graph can be divided in four segments:
OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.
AB: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>
BC: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.
CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.
From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:
Segment AB
