Answer:
W = 222 N.
Explanation:
The qiestion says" If the acceleration of gravity on the surface of the planet Mercury is 3.7 m / s2, then what would be the weight of a person with mass 60 kg on its surface?
"
Mass of the person, m = 60 kg
The acceleration due to gravity on the surface of gravity is 3.7 m/s²
We need to find the weight of a person on the surface of Mercury.
Weight of an object is given by :
W = mg
So,
W = 60 kg × 3.7 m/s²
W = 222 N
Hence, the person will weigh 2222 N on the surface of Moon.
Technician A and B are correct . Because according to technician A, the cause written on the repair order is a diagnosis. Here, by diagnosis, he means that the problem is identified after examining the device and hence the judgement is made.
And according to B, you have to write the cause of the problems in the device that have been identified and the concern measures, which is also kind of diagnosis.
So, option D is correct.
Answer:
Object B has greater density
desity A=20/10=2 g cm^-3 . density B=50/10=5 g cm^-3
the object that has greater mass has the greater density because the volume of the those two objects are same
Answer:
i. The pressure of due to the water, <em>P</em>, is given according to the following equation;
P = ρ·g·h
Where;
ρ = The density of the water (a constant) = 997 kg/m³
g = The acceleration due to gravity = 9.81 m/s²
h = The height of the water (minimum h = h₁, maximum h = h₂)
The pressure is directly proportional to the water height, and we have;
The pressure, <em>P</em>, will be maximum when the water height, <em>h</em>, is maximum or h = h₂, which is the level DC
ii. The thrust = The force acting on the body = Pressure × Area
The maximum areas exposed to the water are on side AB and DC
However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC
Explanation:
