I NEED HELP WITH THE LAST QUESTION ASAP D':

Acceleration is defined as the change in velocity divided by

denpristay [2]

Answer:

Time elapsed

Explanation:

Acceleration is a vector quantity. It is defined as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time elapsed

Acceleration is measured in meters per second squared (m/s^2). It must be noticed that acceleration is a vector, so it also has a direction. In particular:

- when acceleration is negative, it means that the object is slowing down, so acceleration is in opposite direction to the velocity

- when acceleration is positive, it means that the object is speeding up, so acceleration is in the same direction as the velocity

3 0

3 years ago

Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner

stepan [7]

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

7 0

3 years ago

True or false communication is important in healthy relationship

zvonat [6]

Answer:

True

Explanation:

If two partners don't communicate they'll grow apart and it may lead to one or the other cheating.

5 0

3 years ago

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A ball is whirled on the end of a string in a horizontal circle of radius R at speed v. By which one of the following means can

Vera_Pavlovna [14]

Explanation:

When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :

$a=\omega^2R$

Since, $\omega=\dfrac{2\pi }{T}R$

$a=(\dfrac{2\pi}{T})^2R$

T is the time period

R is the radius of the circular path

To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,

R'=6R and T'=2T

So,

$a'=(\dfrac{2\pi}{T'})^2R'$