A billiard ball bangs against a side bumper, giving the ball a speed of 5 m/s. If the

A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a

trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden $d_1=1.1cm$

Speed of water in the hose is $v_1=0.95m/sec$

Number of holes n = 22

Diameter of each holes $d_2=15cm$

According to continuity equation $A_1v_1=A_2v_2$

$d_1^2\times v_1=22\times d_2^2v_2$

$1.1^2\times 0.95=22\times 0.15^2\times v_2$

$v_2=2.322m/sec$

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0

3 years ago

Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de

ahrayia [7]

Answer:

a) El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

$\Delta z = \frac{v_{out}^{2}}{2\cdot g}$

Donde:

$\Delta z$ - Diferencia de altura, medida en metros.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

$v_{out} = \sqrt{2\cdot g \cdot \Delta z}$

Si $g = 9.807\,\frac{m}{s^{2}}$ y $\Delta z = 0.3\,m$ , entonces la rapidez de salida del chorro es:

$v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}$

$v_{out} \approx 2.426\,\frac{m}{s}$

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

$\dot V_{out} = v_{out}\cdot A_{t}$

Donde:

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

$A_{t}$ - Área transversal del orificio, medido en metros cuadrados.

$\dot V_{out}$ - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que $v_{out} = 2.426\,\frac{m}{s}$ y $A_{t} = 5\,cm^{2}$ , el caudal de salida del chorro es:

$\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)$

$\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}$

El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

5 0

3 years ago

Don't mind me I'm trying to get banned

Vilka [71]

Answer is True

Have a good day !

8 0

3 years ago

Some of the largest volcanoes in the solar system are on mars. The most likely explanation is that mars

ivolga24 [154]

Answer: Mars has lower gravity and lower magma buoyancy

Explanation: The gravity in mars is low. Low gravity affect the occurrence of volcanic eruptions. The buoyancy of the magma in Mars is lower and the depth of the magma chamber deeper (Buoyancy is the difference in density between the surrounding crust and the magma ascending for eruption). Olympus Mons which is larger and wider than mount Everest was formed as a result of overcoming of the low gravity and buoyancy by the magma to release enormous volume of magma to the surface.

Another reason why the volcanoes on Mars are so large is because the crust on Mars is static unlike Earth.

4 0

4 years ago

When a voltage difference is applied to a piece of metal wire, a 8-mA current flows through it. If this metal wire is now replac

DanielleElmas [232]

Answer:

Explanation:

The resistance of a wire can be given by the following expression

$R=\frac{\rho\times L}{A }$

where R is resistance , ρ is specific resistance , L is length of wire and A is cross sectional area

specific resistance of metals are almost the same . So in the present case ρ and l are same . Hence the formula becomes

R = k / A where k is a constant .

The diameter of wire becomes two times hence area of cross section becomes 4 times or 4A .

Resistance becomes 1/4 times . Hence if resistance of metal wire is R , resistance of silver wire will be R / 4 .

current = voltage / resistance

In case of metal wire

8 x 10⁻³ = V / R

In case of silver wire

I = V / (R / 4 ) , I is current , V is potential difference .

I = 4 x V/R

= 4 x 8 x 10⁻³ A

= 32 mA.

6 0

4 years ago