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Hunter-Best [27]

2 years ago

12

The specific heat of iron is 0.11 cal (g°C) . A cafeteria fork made of iron has a mass of 20 grams. How much heat energy is need

ed to raise the temperature of this fork from 25°C to 75°C?

1 answer:

Musya8 [376]2 years ago

8 0

It would be, <span>0.275 calories.

Hope this helps.

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

2 years ago

Planets that are larger in diameter, farther from the sun, and less dense than smaller planets in the solar system are known as

Sergio [31]

Gas giants lol I'm love this kinda stuff nothing else just this question

6 0

3 years ago

A skydiver has jumped out of a plane and is falling faster and faster. what forces are present in this situation

kompoz [17]

Gravity and air resistance

i took the test and got 100%

5 0

3 years ago

Dust particles in air have a typical mass of 5.0 x 10-16 kg. They undergo irregular motion due to collisions with air molecules.

Mars2501 [29]

Answer:

Rms speed of the particle will be $38.68\times 10^8m/sec$

Explanation:

We have given mass of the air particle $m=5\times 10^{-16}kg$

Gas constant R = 8.314 J/mol-K

Temperature is given T = $27^{\circ}C=273+27=300K$

We have to find the root mean square speed of the particle

Which is given by $v_{rms}=\sqrt{\frac{3RT}{m}}=\sqrt{\frac{3\times 8.314\times 300}{5\times 10^{-16}}}=38.68\times 10^8m/sec$

So rms speed of the particle will be $38.68\times 10^8m/sec$

5 0

3 years ago

If a 6.1 A resistive load is connected by 2-conductor stranded 2-AWG copper wire to a source voltage of 125.2 V that is 358 ft a

polet [3.4K]

Answer:

124.86 V

Explanation:

We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft

1 ft = 0.3048 m

358 ft = 109.12 m

The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m

The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²

Resistivity of wire (ρ) = 0.0171 Ω.mm²/m

The resistance of the wire = $\frac{\rho A}{l}=\frac{0.0171*109.12 }{33.6} =0.056\ ohm$

The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V

The voltage at end = 125.2 - 0.34 = 124.86 V

3 0

2 years ago