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2 years ago

A runner goes around a 1 mile track if they end up at their starting position what is their dosplacement

Physics

1 answer:

mario62 [17]2 years ago

8 0

Answer:

The displacement is zero

Explanation:

Displacement is a vector that connects the starting position of motion of an object to the final position. It is a vector quantity, so it has both magnitude and direction.

Viceversa, distance is a scalar quantity that measures the total length of the path covered by the object during its motion.

In this problem, we are interested in the displacement.

We are told that the runner goes around a 1 mile track, and he ends at his starting position: this means that his final position is equal to the starting position, therefore the displacement is zero.

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Water ﬂows through a horizontal nozzle in steady ﬂow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2

Dvinal [7]

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

Q = A v

v = Q / A

The area of a circle is

A = π r² = π d² / 4

Let's look at the speeds at each point

v₁ = Q / A₁ = Q 4 /π d₁²

v₁ = 10 4 /π 0.5²

v₁ = 50.93 m / s

v₂ = Q / A₂

v₂ = 10 4 /π 0.25²

v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

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

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

P₁ = P2 + ½ rho (v₂² - v₁²)

P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

P₁ = 1.013 10⁵ + 2.205 10⁷

P₁ = 2.215 10⁷ Pa

la definicion de presion es

P₁ = F₁/A₁

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F₁ = 2.215 10⁷ pi 0.5²/4

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3 years ago

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1. When you have different masses for each sphere, how does the force that the larger mass sphere exerts on the smaller mass sph

aleksandrvk [35]

1) The forces are equal (Newton's third law of motion)

2) The force between the spheres will quadruple

3) The force of gravity exerted by the notebook on you is negligible

Explanation:

In this part of the problem, we want to compare the gravitational force exerted by the larger mass sphere on the smaller mass sphere to the force exerted by the smaller mass sphere to the larger mass sphere.

We can do this by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In this problem, we can identify the larger mass sphere as object A and the smaller mass sphere as object B. This law tells us that the two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the larger mass sphere on the smaller mass sphere is equal to the force exerted by the smaller mass sphere to the larger mass sphere.

The magnitude of the gravitational force between the two spheres is given by

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

$m_1, m_2$ are the masses of the two spheres

r is the separation between the two spheres

In this problem, we are asked to find what happens when the distance between the spheres is halved, therefore when the new distance is

$r'=\frac{r}{2}$

Substituting into the equation, we find

$F'=G\frac{m_1 m_2}{r'^2}=G\frac{m_1 m_2}{(r/2)^2}=4(\frac{Gm_1 m_2}{r^2})=4F$

So, the force between the two spheres will quadruple.

We can give an estimate for the gravitational force exerted by your notebook on you.

As we said, the magnitude of the gravitational force is

$F=G\frac{m_1 m_2}{r^2}$

Where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

Let's estimate the following:

$m_1 = 60 kg$ is your mass

$m_2 = 2 kg$ is the mass of the notebook

$r=1 m$ , assuming the notebook is at 1 metre from you

Substituting,

$F=(6.67\cdot 10^{-11})\frac{(60)(2)}{1^2}=8.0\cdot 10^{-9} N$

We see that this force has an extremely small value: therefore, it is almost negligible in daily life, where other much stronger forces act on you.

Learn more about gravity:

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