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Tresset [83]
3 years ago
14

Which one of the following is NOT a simple machine?

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
6 0
It is a hammer because hammers are not examples of a simple machine
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I have one reason the reaction take place faster because the molecules are going at a faster pace because the temperature is rising 
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All isotopes of a particular element have the same atomic number. How then do the isotopes of a particular element differ?
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Isotopes of an element will contain the same number of protons and electrons but will differ in the number of neutrons they contain. In other words, isotopes have the same atomic number because they are the same element but have a different atomic mass because they contain a different number of neutrons
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4 years ago
A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}

Differentiate both sides with respect to d.

\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

\frac{d^{2}A}{dt^{2}}= + ve

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0
3 years ago
Prompt
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Answer:

1.bandages

2.cottons

3.scissor

4.medicines

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3 years ago
A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).
yawa3891 [41]

To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.

By definition the oscillation frequency is defined as

f = n\frac{v}{2L}

Where

v = speed of sound

L = Length of the pipe

n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)

Re-arrange to find L,

f = n\frac{v}{2L}\\L = \frac{nv}{2f}

The radius between the two frequencies would be 4 to 5,

\frac{528Hz}{660Hz}= \frac{4}{5}

4:5

Therefore the frequencies are in the ratio of natural numbers.  That is

4f = 528\\f = \frac{528}{4}\\f = 132Hz

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m

Therefore the length of the pipe is 1.3m

For the second harmonic n=2, then

L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m

Therefore the length of the pipe in the second harmonic is 2.6m

7 0
3 years ago
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