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3 years ago

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Assume the existence of an abstract class named DesktopComponent with the following: a void (abstract) method, onClicked, that a

ccepts no parameters. a (private) string named type, describing the sort of Desktop component (e.g. window, icon, taskbar, etc). a constructor accepting a string that is used to initialize the type instance variable Write the definition of a subclass, named Window with the following: A. a constructor that invokes the DesktopComponent constructor passing it that value "Window" for the type. B. an onClicked method that prints out the message "Window selected" to System.out

1 answer:

Natalka [10]3 years ago

8 0

Answer:

public class Window extends DesktopComponent {

public Window() {super("Window");}

public void onClicked() {System.out.println("Window selected");}

}

Explanation:

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A 185 g block is pressed against a spring of force constant 1.60 kN/m until the block compresses the spring 10.0 cm. The spring

denis-greek [22]

Answer:

d = 5.10 m

Explanation:

As we know that here on the plane of the inclined there is no frictional force

So in these cases we can say that total mechanical energy will always remains conserved

so here we can say that

spring potential energy = gravitational potential energy of the block

as we know from the formula

$\frac{1}{2}kx^2 = mgh$

now plug in the values in it

$\frac{1}{2}(1.60 \times 10^3)(0.10)^2 = (0.185)(9.81)h$

$8 = 1.81 h$

$h = 4.42 m$

now as we know that the angle of inclination is 60 degree and height raised is 4.42 m

so here maximum distance moved along the inclined plane will be

$\frac{h}{d} = sin60$

$d = \frac{h}{sin60}$

$d = \frac{4.42}{sin60} = 5.10 m$

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3 years ago

You push a cart with mass 15 kg forward, giving it an acceleration of 3 m/s2

NARA [144]

Answer:

45 N

Explanation:

F= ma (ie force is found by multiplying the mass of the object by its accerelation)

thus, F = 15 X 3 = 45 N

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3 years ago

Complete the sentences below using the words below, you may use each word more than once

satela [25.4K]

Decreases, stays the same, increases. The volume decreases because as air is cooled, the individual molecules collectively possess less kinetic energy and the distances between them decrease, thus leading to a decrease in the volume they occupy at a certain pressure (please note that my answer only holds under constant pressure; air, as a gas, doesn't actually have a definite volume). The mass stays the same because physical processes do not create or destroy matter. The law of conservation of mass is obeyed. You're only cooling the air, not adding more air molecules. The density decreases because as the volume decreases and mass stays the same, you have the same mass occupying a smaller volume. Density is mass divided by volume, so as mass is held constant and volume decreases, density increases.

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3 years ago

Which of the following paths correctly displays the transfer of energy for sound?

galben [10]

Energy source to detection to medium

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3 years ago

Read 2 more answers

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

Nina [5.8K]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

3 0

4 years ago