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Setler [38]
3 years ago
15

4 what is the difference between an array's size declarator and a subscript?

Physics
1 answer:
Ber [7]3 years ago
7 0
What is the difference between<span> a</span>size declarator<span> and a </span>subscript<span>? The </span>size declarator<span> is ... When writing a function that accepts a two-dimensional </span>array<span> as an argument, which </span>size declarator<span> must you provide in the parameter </span>for<span> the</span>array<span>? The second size ...</span>
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PLEASE The weight of a girl  is  600 N and the area of her one foot is        50 cm². What will be the pressure exerted by her o
Rus_ich [418]

Answer:

pressure=force/area

p=600/50

p=12

so 12Nm^-1

4 0
3 years ago
While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.
lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

6 0
3 years ago
PLEASE PLEASE HELP!!!Answer the following questions
adoni [48]

Answer:

u have to stop

slow down

move forward

6 0
3 years ago
The left end of a rod of length and rotational inertia is attached to a frictionless horizontal surface by a frictionless pivot,
natima [27]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

6 0
3 years ago
How is the q of an rlc parallel resonant circuit calculated?
notka56 [123]

Answer:

It is calculated by dividing Resistance, R, by Inductive reactance, XL.

Explanation:

Q is called the Q factor of a resonance circuit. In a parallel resonance circuit, it is calculated by finding the ratio of the power stored in the circuit to the power distributed in the circuit. It is a way of measuring the quality of a circuit or how effective the circuit is.

Q factor is the inverse in the resonance series circuit.

Q factor of a resonance parallel circuit,

<h3>Q = R/XL</h3>

R = Resistance

XL = Inductive reactance

3 0
3 years ago
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