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3 years ago

6

The temperature of a 2.0 kg block increases by 5 0c when 2,000 j of thermal energy are added to the block. What is the block's s

pecific heat?

1 answer:

FinnZ [79.3K]3 years ago

7 0

Mass of the block = 2 kg

Temperature is increased by ΔT = 50 °C

Energy consumed by the block in raising the temperature = 2000 Joules

Let the specific heat capacity of the block be "c".

Formula to be used:

Heat gained = m × c × ΔT

Plugging the values:

2000 = 2 × c × 50

$c = \frac{2000}{2\times 50}$

c = 20 J/ kg °C

Hence, the specific heat capacity of the block is: 20 J/ kg °C

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A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp

VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir $T_L=281K$

We know that efficiency of Carnot engine is given by

$\eta =1-\frac{T_L}{T_H}$

$0.26 =1-\frac{281}{T_H}$

$T_H=379.72K$

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so $T_H=379.72K$

So $0.35=1-\frac{T_L}{379.72}$

$T_L=246.818K$

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0

3 years ago

Which of the following is fact-based science rather than part of a personal belief system?

marusya05 [52]

Since there are no choices, then this question calls for open-ended answers. Facts-based science must have proven underlying laws that support inferences such as Coulomb's Law, Kinetic Theory of Matter and many more. On the other hand, examples of science that focus on personal belief is philosophy. This depends on the perspective of known philosophers. An example would be Sigmund Freud who proposed the theory of 3 personalities. Although it is more on personal beliefs, this is used as a foundation in the study of psychology.

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2 years ago

Paying off your debts will most likely

g100num [7]

I think the answer is d

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2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago

A block is released from rest, at a height h, and allowed to slide down an inclined plane. There is friction on the plane. At th

GREYUIT [131]

Here the block has two work done on it

1. Work done by gravity

2. Work done by friction force

So here it start from height "h" and then again raise to height hA after compressing the spring

So work done by the gravity is given as

$W_g = m_A g(h - h_A)$

Now work done by the friction force is to be calculated by finding total path length because friction force is a non conservative force and its work depends on total path

$W_f = -(\mu m_A g cos\theta)(\frac{h}{sin\theta} + \frac{h_A}{sin\theta})$

$W_f = -\mu m_A g cot\theta(h + h_A)$

Total work done on it

$W = m_A g(h - h_A) - \mu m_A g cot\theta(h + h_A)$

So answer will be

None of these

7 0

3 years ago

Read 2 more answers