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2 years ago

5

Given the following specific heat capacities, which material was have the largest change in temperature if 10 grams of each subs

tance absorbs 100 calories of heat?
Substance Heat Capacity - Cal / (g oC)

aluminum 0.22

copper 0.093

lead 0.0305

silver 0.056

2 answers:

Ilya [14]2 years ago

5 0

Answer:

Explanation:

Comment

You could calculate it out by assuming the same starting temperature for each substance. (You have to assume that the substances do start at the same temperature anyway).

That's like shooting 12 with 2 dice. It can be done, but aiming for a more common number is a better idea.

Same with this question.

You should just develop a rule. The rule will look like this

The greater the heat capacity the (higher or lower) the change in temperature.

The greater the heat capacity the lower the change in temperature

That's not your question. You want to know which substance will have the greatest temperature change given their heat capacities.

Answer

lead. It has the smallest heat capacity and therefore it's temperature change will be the greatest.

ElenaW [278]2 years ago

3 0

Find the relationship

$\\ \rm\Rrightarrow Q=mc\Delta T$

$\\ \rm\Rrightarrow 100=10c\Delta T$

$\\ \rm\Rrightarrow 10=c\Delta T$

$\\ \rm\Rrightarrow \dfrac{10}{c}=\Delta T$

$\\ \rm\Rrightarrow \Delta T\propto \dfrac{1}{c}$

So lower the c higher ∆T

Lead has the largest change in T as it has least capacity

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A 0.50 kg mass is attached to a string 1.0 meter long and moves in a horizontal circle at a rate of 0.5 seconds per revolution.

vesna_86 [32]

(1) First compute the linear speed of the mass. If it completes 1 revolution in 0.5 seconds, then the mass traverses a distance of 2<em>π</em> (1.0 m) ≈ 2<em>π</em> m (the circumference of the circular path), so that its speed is

<em>v</em> = (1/0.5 rev/s) • (2<em>π</em> m/rev) = 4<em>π</em> m/s ≈ 12.57 m/s

Then the centripetal acceleration <em>a</em> is

<em>a</em> = <em>v</em>² / (1.0 m) = 16<em>π</em>² m/s² ≈ 160 m/s²

(where <em>r</em> is the path's radius).

(2) By Newton's second law, the tension in the string is <em>T</em> such that

<em>T</em> = (0.50 kg) <em>a</em> = 8<em>π</em>² N ≈ 79 N

7 0

3 years ago

The rate (in liters per minute) at which water drains from a tank is recorded at half-minute intervals. Use the average of the l

lana [24]

Answer:

see explanation

Explanation:

You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:

See the attached table.

From the left we have:

r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min

From the right we have:

r = 1/2 (48 + 46 +44 + 42 + 40 + 38) = 129 L/min.

And this should be the correct answer. Watch your chart and replace if it's neccesary.

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3 years ago

Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550

Arisa [49]

Answer:

q_poly = 14.55 KJ/kg

Explanation:

Given:

Initial State:

P_i = 550 KPa

T_i = 400 K

Final State:

T_f = 350 K

Constants:

R = 0.189 KJ/kgK

k = 1.289 = c_p / c_v

n = 1.2 (poly-tropic index)

Find:

Determine the heat transfer per kg in the process.

Solution:

-The heat transfer per kg of poly-tropic process is given by the expression:

q_poly = w_poly*(k - n)/(k-1)

- Evaluate w_poly:

w_poly = R*(T_f - T_i)/(1-n)

w_poly = 0.189*(350 - 400)/(1-1.2)

w_poly = 47.25 KJ/kg

-Hence,

q_poly = 47.25*(1.289 - 1.2)/(1.289-1)

q_poly = 14.55 KJ/kg

4 0

3 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

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3 years ago

The acceleration of a particle is given by a = 3t – 4, where a is in meters per second squared and t is in seconds. Determine th

Genrish500 [490]

Answer:

$v=0.04m/s\\$

$s=-28.592m\\$

Explanation:

$a = 3t-4$

$v(t)=\int\limits^t_0 {a(t)} \, dt =3/2*t^{2}-4t+v_0\\$

if t=3.6s and initial velocity, v0, is -5m/s

$v=0.04m/s\\$

$s(t)=\int\limits^t_0 {v(t)} \, dt =1/2*t^{3}-2t^{2}+v_0*t+s_0\\$

if t=3.6s and the initial displacement, s0, is -8m:

$s=-28.592m\\$

5 0

3 years ago