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Alinara [238K]
3 years ago
6

Before using a string in a comparison, you can use either the To Upper method or the To Lower method to convert the string to up

per case or lower case, respectively, and then use the converted string in the comparison.1. True2. False

Physics
1 answer:
diamong [38]3 years ago
3 0

Answer:

True, check attachment for code

Explanation:

To convert java strings of text to upper or lower case, we can use and inbuilt methods To Uppercase and To lower case.

The first two lines of code will set up a String variable to hold the text "text to change", and then we print it out.

The third line sets of a second String variable called result.

The fourth line is where the conversion is done.

We can compare the string

We can compare one string to another. (When comparing, Java will use the hexadecimal values rather than the letters themselves.) For example, if we wanted to compare the word "Fat" with the word "App" to see which should come first, you can use an inbuilt string method called compareTo.

Check attachment for the code

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PLEASE HELP!!!! OFFERING 99 POINTS!!
Nitella [24]

The Gravitational PE (U) depends on three things: the object’s mass (m), its height (h), and gravitational acceleration (g), which is 9.81 m/s^2 on Earth’s surface.

so U = mgh = 9.81mh on earth

mass of the car = 50.0 grams = 0.05kg

height, h:

Hill 1 = 90.0 cm = 0.9m,

Hill 2 = 65.0 cm = 0.65m,

Hill 3 = 20.0 cm = 0.2m

substitute into eqn U = mgh

U @ top of Hill 1 = 0.05*9.81*0.9 = 0.4415J

U @ top of Hill 2 = 0.05*9.81*0.65 = 0.3188J

U @ top of Hill 3 = 0.05*9.81*0.2 = 0.0981J

difference in Gravitational Potential Energy from the top of Hill 1 to the top of Hill 3 = 0.4415 - 0.0981

= 0.3434J where J is the unit for energy, Joules


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3 years ago
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Which is a characteristic of atoms?
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2 years ago
PLZ answer asp The image shows the combustion of methane. Which equation represents this chemical reaction?
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A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radi
svetlana [45]

Answer:

E=0 at r < R;

E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}} at 2R > r > R;

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}} at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon} (integral over a closed surface)

where,

E = Electric field

Q_{enclosed} = charged enclosed within the closed surface

\epsilon = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

Q_{enclosed} = 0 and hence E = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

Q_{enclosed} = Q,

therefore,

E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to 4\pi r^{2})

or, E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}

at r >= 2R

Q_{enclosed} = 2Q

Hence, by similar calculations, we get,

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}

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All objects in orbit must follow the path of an ellipse (one of Keplers laws)
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