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2 years ago

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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]2 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

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A weight lifter lifts a 345 N set of weights from ground level to a position over his head, a vertical distance of 1.89 m. How m

Pavlova-9 [17]

Answer:652.05 J

Explanation:

Given

Weight of lifter $W=345 N$

vertical distance move $h=1.89 m$

Work done in lifting the weight is equal change in Potential Energy of weight

Change in Potential Energy $=m g h$

$\Delta PE=345 \times 1.89=652.05 J$

therefore work done is equal to $652.05 J$

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2 years ago

In a closed system, what happens to the total energy of the system as energy conversion takes place?

scoundrel [369]

Answer:

In a closed system, the total energy is conserved or remains the same as energy transformations take place.

Explanation:

The law of conservation of energy states that energy cannot be created or destroyed but can be transformed from one form to another.

This law of conservation of energy applies only to a closed system. A closed system is a system which does not exchange energy with its surroundings. All forms of energy conversions occurring within a closed system does not result in an increase or decrease of the total energy of the system, rather, energy remains constant. For example, the universe is a closed system in that all forms of energy conversions occurs within it and energy is not exchanged with an external environment. However, the earth is not a closed system as some of the energy it receives from the sun can be radiated out into space. Since it's an open system, its total energy can change.

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1 year ago

A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra

son4ous [18]

Answer:

The answer is $v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}$ .

Explanation:

From the law of gravity,

$F = G \frac{Mm}{r^2}$

considering F as a conservative force, $F = - \nabla U$ ,

the general expression for gravitational potential energy is

$U = -G \frac{Mm}{r}$ ,

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

$g = -G \frac{M}{r^2}$ ,

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

$U = g M_e R$ ,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

$U_p = g M_e(R- r_s)$ ,

then, the kinetik energy when the rock hits the surface is

$U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s)$ ,

so

$v = \sqrt{2g(R-r_s)}$

where g is the gravitational acceleration generated by the Sun at R,

$g = G \frac{M_s}{R^2}$ .

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2 years ago

In which one or more of the following is the earth assumed to be the center of the universe?

sladkih [1.3K]

Can you put the answers, if there are any?
Once you do, Ill respond with a answer asap!
:)

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2 years ago

Read 2 more answers

E14. A ball rolls off a table with a horizontal velocity of 5 m/s. If

Shkiper50 [21]

a) Vertical velocity: 5.9 m/s

b) Horizontal velocity: 5 m/s

Explanation:

a)

The motion of the ball is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction)

- A uniformly accelerated motion (constant acceleration) along the vertical direction

Here we want to find the vertical component of the ball's velocity. This can be done by using the suvat equation for the vertical motion:

$v_y = u_y +gt$

where:

$v_y$ is the vertical velocity at time t

$u_y=0$ is the initial vertical velocity (zero because the ball has been thrown horizontally)

$g=10 m/s^2$ is the acceleration of gravity (here we take downward as positive direction)

Substituting t = 0.6 s, which is the total time of flight, we find the vertical velocity of the ball just before it hits the ground:

$v_y=0+(9.8)(0.6)=5.9 m/s$

b)

The motion along the vertical direction is an accelerated motion, because there is a force (the force of gravity) acting on the ball and that it causes an acceleration in the ball.

However, there are no forces acting in the horizontal direction on the ball (if we neglect the air resistance): this means that the acceleration of the ball in the horizontal direction is zero.

As a consequence, this also means that the horizontal component of the ball's velocity is constant during the motion.

Since the ball was thrown from the table with an initial horizontal velocity of 5 m/s, this means that the horizontal velocity of the ball just before it hits the floor is still

$v_x = 5 m/s$

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3 years ago