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ch4aika [34]
3 years ago
10

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:
Burka [1]3 years ago
5 0

Answer:

True

Explanation:

If there's no preference over the string case (upper case or lower case), one can convert both strings to upper case or to lowercase and then compare the converted strings to test if they're equal or not.

An Illustration is

string a = "Boy"

string b = 'bOy"

if(a.ToUpper() == b.ToUpper() || a.ToLower() == b.ToLower())

{

Print "Equal Strings"

}

else

{

Print "Strings are not equal";

}

The above will first convert both strings and then compare.

Since they are the same (after conversion), the statement "Equal Strings" will be printed, without the quotes

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An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid
serg [7]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is r_x=140 km

Distance of astronaut From asteroid Y is r_y=581 km

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

F_x=F_y

\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}

cancel out the common terms we get

\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}

\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2

\frac{M_x}{M_y}=(\frac{140}{581})^2

\frac{M_x}{M_y}=0.05806\approx 0.0581

8 0
3 years ago
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
3 years ago
A ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10
irina [24]
First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.
5 0
3 years ago
Read 2 more answers
What is current electricity? write down its use in our daily like ​.
Karolina [17]

Answer: Current electricity is a form of electricity in which charges constantly flow. Current electricity is dynamic while static electricity, as the name suggests, is static. How does current electricity work? The steady flow of electrons is termed as current electricity. Uses of Electricity in Household Starting from toaster to refrigerator, microwave, washing machine, dishwasher, electrical chimney, and many more appliances which are simple to use and made for the convenience of day to day activities use electricity to function.

8 0
3 years ago
) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi
natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0
3 years ago
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