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2 years ago

PLS HELP ME ASAPPPP. 50 POINTSSS

Physics

1 answer:

dybincka [34]2 years ago

4 0

Answer:

1.8 kj

Explanation:

Explanation:

A substance's specific heat tells you how much heat is required to increase the mass of

1 g

of that substance by

∘

The equation that establishes a rel;ationship between heat absorbed and change in temperature looks like this

⋅

, where

- heat absorbed

- the mass of the sample

- the specific heat of the substance

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

You have all the information needed to find the amount of heat required to increase the temperature of your sample of mercury by that many degrees Celsius, so just rearange the above equation and solve for

250.0

⋅

0.14

∘

⋅

(

−

)

∘

1820 J

I'll leave the answer rounded to two sig figs and expressed in kilojoules

1.8 kJ

Answer Explanation:

A substance's specific heat tells you how much heat is required to increase the mass of

1 g

of that substance by

∘

The equation that establishes a rel;ationship between heat absorbed and change in temperature looks like this

⋅

, where

- heat absorbed

- the mass of the sample

- the specific heat of the substance

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

250.0

⋅

0.14

∘

⋅

(

−

)

∘

1820 J

I'll leave the answer rounded to two sig figs and expressed in kilojoules

1.8 kJ

Answer link

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A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

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Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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Answer:

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The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

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Fr = Fr

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