Which terms describes elements with different atomic masses due to varying numbers of neutrons?

A garden hose with a diameter of 0.64 in has water flowing in it with a speed of 0.46 m/s and a pressure of 1.9 atmospheres. At

STALIN [3.7K]

Answer:

(a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

Explanation:

Given that,

Nozzle diameter = 0.25 in = 0.00635 m

Hose pipe diameter = 0.64 in = 0.016256 m

Pressure = 1.9 atm =192518 Pa

(a). We need to calculate the speed of the water in the nozzle

Flow Speed at the inlet pipe will be given by using Continuity Equation

$Q_{1}=Q_{2}$

$v_{1}A_{1}=v_{2}A_{2}$

$v_{1}=v_{2}\times(\dfrac{A_{2}}{A_{1}})$

Where, A = area of pipe

$A=\pi\times \dfrac{d^2}{4}$

$v_{1}=v_{2}\times(\dfrac{d_{2}^2}{d_{1}^2})$

Put the value into the formula

$v_{1}=0.46\times\dfrac{(0.016256)^2}{(0.00635)^2}$

$v_{1}=3.014\ m/s$

The speed of the water in the nozzle is 3.014 m/s.

(b). We need to calculate the pressure in the nozzle

Using Bernoulli's Theorem,

$P_{1}+\dfrac{1}{2}\rho\times v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2+\rho gh_{2}$

Where, $h_{1}=h_{2}$

$P_{1}+\dfrac{1}{2}\rho\times v_{1}^2=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2$

$P_{1}=P_{2}+\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)$

Put the value into the formula

$P_{1}=192518 +\dfrac{1}{2}\times1000\times((0.46)^2-(3.014)^2)$

$P_{1}=188081.702\ Pa$

$P=1.86\ atm$

Hence, (a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

7 0

3 years ago

Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.

MA_775_DIABLO [31]

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>

It is given that, the infinite well having a width of 0.050 mm.
We have the expression for energy of electron in an infinite well as,

$E_n=\frac{n^2h^2}{8mL^2}$

where;

$m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1$

Thus, the lowest energy of electron in an infinite well is,

$E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J$

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

Learn more about the infinite well here:

brainly.com/question/20317353

#SPJ4

7 0

1 year ago

If you fall from a building onto a net which extends the time of impact by 10 times, what happens to the force you experience?

zubka84 [21]

Answer:

A larger impulse. A 1-kg ball has twice as much speed as a 10-kg ball.

Explanation:

6 0

3 years ago

A ball of mass 5 kg attached to a string is swung in a horizontal circle of radius 0.5 m. If the tension in the string is 10 N,

kirill115 [55]

Answer:

0 J

Explanation:

given,

mass of the ball = 5 kg

radius of the horizontal circle = 0.5 m

tension in the string = 10 N

Work done = ?

Work done by the tension in the circular path will be equal to zero.

This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.

so, work done = F s cos θ

θ = 90°,

work done = F s cos 90° ∵ cos 90° = 0

Work done = 0 J

8 0

3 years ago

The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3

viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

$x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;$

$\Rightarrow mg\sin{\theta} = F$

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at $\theta.$ Solving for the angle, we get

$\sin{\theta} = \dfrac{F}{mg}$

or

$\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)$

$\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]$

$\;\;\;=42.9°$

6 0

2 years ago