Answer:
buoyant force on the block due to the water= 10 N
Explanation:
We know that
buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.
Given:
A block of metal weighs 40 N in air and 30 N in water.
F_B = 40-30= 10 N
therefore, buoyant force on the block due to the water= 10 N
the sentence with leeward side and the sentence that has windward side both have errors.
<span>The answer of these two problems are :
A) a = (F-W)/m = (100-500)/500/g = 9.8*500/500 = +9.8
m/sec^2
B) a1 = +9.8-g = 0</span>
This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
(a) Both the girl and the boy have the same nonzero angular displacement.
Explanation:
The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by

where
is the angular speed
t is the time interval
For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of
is the same for the boy and the girl.
Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.
(b) The girl has greater linear speed.
Explanation:
The linear (tangential) speed of a point along the merry-go-round is given by

where
is the angular speed
r is the distance of the point from the centre of the merry-go-round
In this problem, the girl is near the outer edge, while the boy is closer to the centre: since the value of
is the same for both, this means that the value of r is larger for the girl, so the girl will also have a greater linear speed.