Ask question

Ask question

All categories

3 years ago

15

A garden hose with an internal diameter of 2.9 cm is connected to a (stationary) lawn sprinkler that consists merely of a contai

ner with 24 holes, each 0.36 cm in diameter. If the water in the hose has a speed of 0.98 m/s, at what speed does it leave the sprinkler holes?

1 answer:

Alecsey [184]3 years ago

3 0

Answer:

Water leaves the holes with a speed of 2.65 m/sec

Explanation:

It is given internal diameter of garden hose = 2.9 cm

So internal radius $r_1=\frac{2.9}{2}=1.45cm$

Number of holes n = 24

Diameter of holes d = 0.36 cm

So radius of holes $r_2=\frac{0.36}{2}=0.18cm$

Velocity of water in hose $v_1=0.98m/sec$

According to continuity equation

$A_1v_1=nA_2v_2$

$\pi r_1^2\times 0.98=24\times \pi r_2^2\times \times v_2$

$1.45^2\times 0.98=24\times 0.18^2\times \times v_2$

$v_2=2.65m/sec$

Therefore water leaves the holes with a speed of 2.65 m/sec

You might be interested in

__________ and __________ both agreed that childhood experiences play an important role in the development of oneself.

oksano4ka [1.4K]

The following answers in the blank could be Sigmund Freud who found the psychoanalysis and Karen Horney who is known to be a psychoanalyst. Both of them has the idea that childhood experiences is necessary and plays the important role in the individual as he or she grows.

8 0

3 years ago

Read 2 more answers

How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)

mafiozo [28]

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

5 0

3 years ago

There are some points on a standing

bija089 [108]

Answer:

Nodes.

Explanation:

Nodes are a point that are on a standing wave that never move.

7 0

3 years ago

Read 2 more answers

A car accelerates at a rate of 13m/s^2[S]. If the car's initial velocity is 120km/h[N]. What will its final velocity be in m/s,

Delvig [45]

Answer:

the final velocity of the car is 59.33 m/s [N]

Explanation:

Given;

acceleration of the car, a = 13 m/s²

initial velocity of the car, u = 120 km/h = 33.33 m/s

duration of the car motion, t = 2 s

The final velocity of the car in the same direction is calculated as follows;

v = u + at

where;

v is the final velocity of the car

v = 33.33 + (13 x 2)

v = 59.33 m/s [N]

Therefore, the final velocity of the car is 59.33 m/s [N]

6 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

5 0

3 years ago