Answer: The given statement is false.
Explanation:
According to Newton's third law of motion, every action has an equal and opposite reaction. So, when we apply force in one direction on an object then the object also applies a force in the opposite direction.
Hence, it is true that two forces in each pair of forces act in opposite directions.
For example, when we push a wooden box of 20 kg in the forward direction then the box will also apply a force in the opposite direction.
But the statement two forces in each pair can either both act on the same body or they can act on different bodies is false.
Answer:
option (c)
Explanation:
When an object thrown upwards, the value of acceleration acting on the object is acceleration due to gravity which is always acting towards the earth.
As it falls downwards, the acceleration is again equal to the acceleration due to gravity.
So, the ball's acceleration is constant.
Answer:
Mass = 64,870,000,000 kilograms
Explanation:
Given the following data;
Density = 998 kg/m³
Volume = 6,500,000 m³
To find the mass of water in the reservoir;
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
Density = mass/volume
Making mass the subject of formula, we have;
Mass = density * volume
Mass = 998 * 6,500,000
Mass = 64,870,000,000 kilograms
Answer:
P₁ = 2.3506 10⁵ Pa
Explanation:
For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
A₁ v₁ = A₂ v₂
Let's look for the areas
r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm
r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm
A₁ = π r₁²
A₁ = π 1.125²
A₁ = 3,976 cm²
A₂ = π r₂²
A₂ = π 0.1²
A₂ = 0.0452 cm²
Now with the continuity equation we can look for the speed of water inside the hose
v₁ = v₂ A₂ / A₁
v₁ = 11.2 0.0452 / 3.976
v₁ = 0.1273 m / s
Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)
P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂
P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂
Let's calculate
P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25
P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴
P₁ = 2.3506 10⁵ Pa
