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3 years ago

This ball is technology too! It can be rolled, kicked, or thrown. Is that a need or a want?

Physics

2 answers:

Inga [223]3 years ago

7 0

Answer:

want

Explanation:

people always want the new thing especially if its technology

Juli2301 [7.4K]3 years ago

5 0

Wantttttrtttt............

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An aluminum cup contains 225 g of water and a 40 g copper stirrer, all at 27°C. A 410 g sample of silver at an initial temperatu

fomenos

Answer:

130.22 g

Explanation:

Parameters given:

Mass of water Mw = 225 g

Mass of stirrer Ms = 40 g

Mass of silver M(S) = 410 g

By applying the law of conservation of energy:

(McCc + MsCs + MwCw)ΔTw = M(S)C(S)ΔT(S)

where Mc = Mass of cup

Cc = Specific heat capacity of aluminium cup = 900 J/gC

Cs = Specific heat capacity of copper stirrer = 387 J/gC

Cw = Specific heat capacity of water = 4186 J/gC

ΔTw = change in temperature of water = 32 - 27 = 5 °C

C(S) = Specific heat capacity of silver = 234 J/gC

ΔT(S) = change in temperature of silver = 88 - 32 = 56 °C

Therefore:

[(Mc * 900) + (40 * 387) + (225 * 4186)] * 5 = 410 * 234 * 56

(900Mc + 957330) * 5 = 5276700

900Mc + 957330 = 5276700 / 5 = 1074528

900Mc = 1074528 - 957330

900Mc = 117198

Mc = 117198/ 900

Mc = 130.22 g

The mass of the cup is 130.22 g.

5 0

3 years ago

Jasmine and her partner want to test to see if there is a relationship between how much of her solar panel is exposed to sunligh

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Any one trial might have been done incorrectly.

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2 years ago

Dr. Man was studying a fossil from a dig in a desert when he noticed that the organism had evidence of fins.

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What organism? Picture?

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2 years ago

One of the harmonic frequencies of tube A with two open ends is 576 Hz. The next-highest harmonic frequency is 648 Hz. (a) What

balu736 [363]

(a) 288 Hz

The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, $f_1$ (first harmonic):

$f_{n+1}-f_n = f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 576 Hz\\f_{n+1}=648 Hz$

So the fundamental frequency is:

$f_1 = 648 Hz-576 Hz=72 Hz$

Now we know that one of the the harmonics is $f_n=216 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+f_1 = 216 Hz+72 Hz=288 Hz$

(b) n=4

The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:

$f_n=n f_1$ (2)

Since we know $f_n = 288 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{f_n}{f_1}=\frac{288 Hz}{72 Hz}=4$

(c) 4445 Hz

For a closed pipe (only one end is open), the situation is a bit different, because only odd harmonics are allowed. This means that the frequency of the nth-harmonic is an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

so, the difference between any two harmonics tube is equal to:

$f_{n+1}-f_n = (2(n+1)+1)f_1-(2n+1)f_1=(2n+3)f_1-(2n+1)f_1=2f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 4699 Hz\\f_{n+1}=4953 Hz$

So, according to (1), the fundamental frequency is equal to half of this difference:

$f_1 = \frac{4953 Hz-4699 Hz}{2}=127 Hz$

Now we know that one of the harmonics is $f_n=4191 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+2f_1 = 4191 Hz+254 Hz=4445 Hz$

(d) n=17

We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

Since we know $f_n = 4445 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{1}{2}(\frac{f_n}{f_1}-1)=\frac{1}{2}(\frac{4445 Hz}{127 Hz}-1)=17$

7 0

3 years ago

A wave with low energy will also have

yKpoI14uk [10]

Low frequency, low amplitude, or both.

4 0

3 years ago