the reading on a mercury barometer at mombasa is 760mms.calculate the pressure at mombasa(density of mercury=1.36x10 power 4kg/m
1 answer:
Answer:
The pressure at Mombasa is 101.396 KPascal.
Explanation:
Given: Reading on barometer = 760 mm
= 0.76 m
Density of mercury = 1.36 x kg/
In this case, the pressure can be expressed as:
P = σhg
Where: σ is the density of mercury, h is height of mercury in the barometer, and g is the acceleration due of gravity.
But, g = 9.81 m/
So that,
P = 1.36 x x 0.76 x 9.81
= 101396.16 Pascal
P = 101.395 KPascal
The pressure at Mombasa is 101.396 KPascal.
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Answer:
a) = 692 N
b) = 932 N
Explanation:
a)
According to newton's second law of motion, acceleration of an object is directly proportional to the net force acting on it. When there is no net force force acting on the body, there is no acceleration. A force is a push or a pull, and the net force ΣF is the total force, or sum of the forces exerted on an object in all directions.
∝ a
= ma
= ma
Given data:
= 800 N
Mass = m = 90 kg
acceleration = a = 1.2 m/s²
= ?
800 - = (90)(1.2)
= 692 N
b)
According to newton's second law of motion,
∝ a
= ma
= ma
Given data:
If we assume the same friction and acceleration between player's feet and ground as calculated in part a
= 692 N
acceleration = a = 1.2 m/s²
We take the equal mass to the total mass of both the players because when the winning player push losing player backward, he exert force on the ground not only due to his mass but also due to the mass of losing player.
Mass = M = m₁ + m₂ = 110 kg + 90 kg
= 200 kg
= ?
- 692 N = (200)(1.2)
= 692 + 240
= 932 N