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Feliz [49]
3 years ago
13

How much energy, in Joules, is needed to raise the temperature from 25.87°C to 33.16°C in a 2.66 kg block of each of the followi

ng materials? Silver has a specific heat of 0.235 J/g·K, iron has a specific heat of 0.449 J/g·K, and titanium has a specific heat of 0.522 J/g·K. (a) silver
Physics
1 answer:
RUDIKE [14]3 years ago
6 0

Answer:

(a) Silver, H = 455.67 J

(b) Iron, H = 870.67 J

(c) titanium, H = 1012.23 J

Explanation:

mass of block, m = 2.66 kg = 266 g

T1 = 25.87° C

T2 = 33.16° C

ΔT = T2 - T1 = 33.16 - 25.87 = 7.29° C

(a) Specific heat of silver, c = 0.235 J/g K

The heat required is given by

H = m x c x ΔT

H = 266 x 0.235 x 7.29 = 455.67 J

(b) Specific heat of iron, c = 0.449 J/g K

The heat required is given by

H = m x c x ΔT

H = 266 x 0.449 x 7.29 = 870.67 J

(c) Specific heat of titanium, c = 0.522 J/g K

The heat required is given by

H = m x c x ΔT

H = 266 x 0.522 x 7.29 = 1012.23 J

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The frequency of rotation of Mars is 0.0000113 Hertz.

<u>Given the following data:</u>

  • Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

<u>Conversion:</u>

1 day = 24 hours

24 hours to minutes = 60 × 24 = 1440 minutes

1440 + 37 = 1477 \; minutes

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1477 minute = X seconds

Cross-multiplying, we have:

X = 60 × 1477

X = 88620 seconds

Now, we can find the frequency of rotation of Mars by using the formula:

Frequency = \frac{1}{Period}\\\\Frequency = \frac{1}{88620}

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Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.

Read more: brainly.com/question/14708169

8 0
2 years ago
An object of mass 30 kg is in free fall in a vacuum where there is no air resistance. Determine the acceleration of the object.
velikii [3]

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7 0
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A transformer consists of 290 primary windings and 824 secondary windings. Part A: If the potential difference across the primar
jasenka [17]

Answer:

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Part 2) Current in secondary windings is 0.53 Amperes

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The potential developed in the primary and secondary winding of a transformer are related as

\frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}

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Vs Voltage of turns in secondary coil

Applying values in the formula we get

\frac{290}{824}=\frac{21.5}{V_{s}}\\\\\therefore V_{s}=21.5\times \frac{824}{290}=61.08V

Part 2)

Using Ohm's law the current is given by

I=\frac{V_{s}}{R}\\\\I=\frac{61.089}{115}=0.53A

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Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm
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now from above equation we have

256 (20.0 rpm) = 160(\omega_2)

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