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Olegator [25]
3 years ago
11

1.)Describe an experience where you encountered a buoyant force and tell what it felt like.

Physics
1 answer:
kap26 [50]3 years ago
8 0

Swimming: Knowing I would not sink made feel safe.

Taking off in an aircraft: I felt heavier.

<u>Explanation:</u>

The buoyant force originates from the weight applied to the item by the liquid. Since the weight increments as the profundity press, the base of an article are constantly bigger than the power on the top - consequently the net upward power.

It follows up on an article inverse to gravity by liquid which is being submerged mostly or totally in the liquid. It contradicts the heaviness of the item. The buoyant force is given by volume dislodged by an item into the thickness of liquid into gravitational quickening.

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Answer:

sifting

Explanation:

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A fly is call inside a car that is moving in the positive direction with constant velocity. The fly flies with a constant veloci
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A. The fly's velocity to the ground is less than the car's velocity to the ground.
4 0
4 years ago
On a bet, you try to remove water from a glass by blowing across the top of a vertical straw immersed in the water. What is the
MArishka [77]

Answer:

       v₂ = 0.56 m / s

Explanation:

This exercise can be done using Bernoulli's equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Where points 1 and 2 are on the surface of the glass and the top of the straw

The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of ​​the straw the velocity of the surface of the vessel is almost zero v₁ = 0

The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m

We substitute in the equation

         P_{atm} + ρ g y₁ = P_{atm} + ½ ρ v₂² + ρ g y₂

         ½ v₂² = g (y₂-y₁)

        v₂ = √ 2 g (y₂-y₁)

Let's calculate

        v₂ = √ (2 9.8 1.6 10⁻²)

       v₂ = 0.56 m / s

5 0
3 years ago
The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
SpyIntel [72]

Answer:

Magnitude of the Frictional force = (mv₀²)/2x₁

Explanation:

For the frictional force to stop the box, it has to produce the deceleration of the box; thereby being the opposing force to the box's motion.

According to Newton's first law of motion

Frictional force = (mass of the box) × (deceleration experienced by the box)

Let the mass of the box be m

Then,

Frictional force = ma

Then we can obtain the deceleration using the equations of motion

v² = u² + 2ax

u = Initial velocity = v₀ m/s

v = Final velocity = 0 m/s (since the box comes to rest at the end)

x = horizontal distance covered = (x₁ - x₀) = x₁ (since x₀ = 0)

a = ?

v² = u² + 2ax

0 = (v₀)² + 2ax₁

2ax₁ = - v₀²

a = - (v₀²)/(2x₁) (minus sign, because it's a deceleration)

Magnitude of the Frictional force = ma = (mv₀²)/2x₁

4 0
3 years ago
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