Omplete the following program: [0.5 X 4 = 2]

Write an expression that evaluates to true if the value of variable lastName is greater than the string Dexter.

gregori [183]

Answer:

lastName.compareTo("Dexter")>0

Explanation:

The expression that evaluates to true if the value of variable lastName is greater than the string Dexter is; lastName.compareTo("Dexter")>0.

4 0

2 years ago

How do i get a used eraser out of a mechanical pencil

timurjin [86]

Answer:

Use a screw or corkscrew. Screw it into the eraser, then pull.

4 0

2 years ago

Read 2 more answers

A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6

amm1812

1) $a_x=4.287+2.772x\\a_y=-5.579+2.772y$

2) 8.418

Explanation:

1)

The two components of the velocity field in x and y for the field in this problem are:

$u=1.85+2.05x+0.656y$

$v=0.754-2.18x-2.05y$

The x-component and y-component of the acceleration field can be found using the following equations:

$a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}$

$a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}$

The derivatives in this problem are:

$\frac{du}{dt}=0$

$\frac{dv}{dt}=0$

$\frac{du}{dx}=2.05$

$\frac{du}{dy}=0.656$

$\frac{dv}{dx}=-2.18$

$\frac{dv}{dy}=-2.05$

Substituting, we find:

$a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x$

And

$a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y$

2)

In this part of the problem, we want to find the acceleration at the point

(x,y) = (-1,5)

So we have

x = -1

y = 5

First of all, we substitute these values of x and y into the expression for the components of the acceleration field:

$a_x=4.287+2.772x\\a_y=-5.579+2.772y$

And so we find:

$a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281$

And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:

$a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418$

4 0

2 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where the

Archy [21]

Answer:

Work transfer is - 97.02 KJ. It means that work is given to the system.

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.

Explanation:

Given that

m= 3 kg

P₁=2 bar

T=T₁=T₂=30 °C

T=303 K

P₂=2.5 bar

PV= Constant

This is the isothermal process .

We know that work for isothermal process given as

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=mRT\ln \dfrac{P_1}{P_2}$

For air

R= 0.287 KJ/Kg.K

Now by putting the values

$W=mRT\ln \dfrac{P_1}{P_2}$

$W=5\times 0.287\times 303\ln \dfrac{2}{2.5}$

W= - 97.02 KJ

So the work transfer is - 97.02 KJ. It means that work is given to the system.

We know that for ideal gas internal energy is the only function of temperature.The change in internal energy ΔU

ΔU = m Cv ΔT

Here ΔT= 0

So

ΔU =0

From first law of thermodynamics

Q= ΔU +W

ΔU = 0

Q= W

Q= - 97.02 KJ

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.

8 0

3 years ago

(a) what is Linear equation (b) Why Laplace's equation is linear

Triss [41]

Answer:

A) Linear Equation -

Linear equation has only one independent variable and when the linear equation plotted on a graph it forms a straight line. It is made up of two expressions equal to each other in a equation. Linear equation graph fits the Y= mx+a ( m=slope).

B) Laplace's equation is linear as it is a second order partial differential equation. So if we put dependent variable in differential equation it always show result in linear.

7 0

2 years ago