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1 year ago

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A 425-g piece of metal at 100°C is dropped into a 100-g aluminum cup containing 500 g of water at 15°C. The final temperature of

the system is 40°C. What is the specific heat of the metal, assuming no heat is exchanged with the surroundings? The specific heat of aluminum is 900 J/kg°C.
a. 2140 J/(kg•K)
b. 3800 J/(kg•K)
c. 3300 J/(kg•K)
d. 1900 J/(kg•K)
e. 4280 J/(kg•K)

1 answer:

yKpoI14uk [10]1 year ago

5 0

The specific heat of the metal, assuming no heat is exchanged with the surroundings is 2140 J/(kg•K).

<h3>Specific heat capacity of the metal</h3>

The specific heat capacity of the metal is determined from the principle of conservation of energy.

energy lost by the metal = energy gained by aluminum + energy gained by water

Q = mcΔθ

where;

m is mass (kg)
c is specific heat capacity
Δθ is change in temperature

0.425c(100 - 40) = 0.1(900)(40 - 15) + 0.5(4186)(40 - 15)

25.5c = 2250 + 52,325

c = 54,575/25.5

c = 2140 J/(kg•K)

Learn more about specific heat capacity here: brainly.com/question/21406849

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8. A spring of spring constant 4 N/m is stretched 0.5 meters. How strong is the restoring force?

pogonyaev

So, the recovery force of the spring is <u>2 N in the opposite direction of the pull</u>.

<h3>Introduction</h3>

Hi ! Here, I will help you about the spring recovery force. <u>The restoring force is the force that opposes the direction of the initial pull of the spring (either when the spring is pulled horizontally or vertically)</u>. The restoring force is strongly influenced by the type of spring (through the spring constant) and the length of the strain that occurs. Negative values in spring restoring force only indicate direction, not value. The equation that applies is as follows:

If the spring is pulled horizontally

$\boxed{\sf{\bold{F = - k \times \Delta x}}}$

With the following condition :

F = recovery force (N)
k = spring constant (N/m)
$\sf{\Delta x}$ = horizontal length change (m)

If the spring is pulled vertically

$\boxed{\sf{\bold{F = - k \times \Delta y}}}$

With the following condition :

F = recovery force (N)
k = spring constant (N/m)
$\sf{\Delta y}$ = vertical length change (m)

<h3>Problem Solving</h3>

We know that :

Assume the spring is pulled horizontally
k = spring constant = 4 N/m
$\sf{\Delta x}$ = horizontal length change = 0.5 m

What was asked :

F = recovery force = ... N

Step by step :

$\sf{F = - k \times \Delta x}$

$\sf{F = - 4 \times 0.5}$

$\boxed{\sf{F = -2 \: N}}$

<h3>Conclusion :</h3>

So, the recovery force of the spring is 2 N in the opposite direction of the pull.

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3 years ago

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Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

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Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

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