All categories

3 years ago

The first law of thermodynamics can be given as ________.

1 answer:

7 0

The first law of thermodynamics can be written as
$\Delta U = Q-W$
where
$\Delta U$ is the variation of internal energy of the system
$Q$ is the amount of heat absorbed by the system
$W$ is the work done by the system on the surrounding.

Using this form, the sign convention for Q and W becomes:
Q > 0 --> heat absorbed by the system (because it increases the internal energy)
Q < 0 --> heat released by the system (because it decreases the internal energy)
W > 0 --> work done by the system (for instance, an expansion: when the system expands, it does work on the surrounding, and so the internal energy decreases, this is why there is a negative sign in the formula Q-W)
W < 0 --> work done by the surrounding on the system (for instance, a compression: when the system is compressed, the surrounding is doing work on the system, and so the internal energy of the system increases)

You might be interested in

Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov

Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

$K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}$

Final kinetic energy is :

$K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}$

Now, fraction of initial kinetic energy loss is :

$Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25$

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0

3 years ago

The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

Law Incorporation [45]

The gravitational force between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

The distance of the telescope from the Earth's center is $r=6940 km=6.94 \cdot 10^6 m$ , the gravitational force is $F=9.21 \cdot 10^4 N$ and the mass of the Earth is $m_1=5.98 \cdot 10^{24} kg$ , therefore we can rearrange the previous equation to find m2, the mass of the telescope:
$m_2 = \frac{Fr^2}{Gm_1}= \frac{(9.21 \cdot 10^4 N)(6.94\cdot 10^6)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})} =11121 kg$

6 0

3 years ago

Read 2 more answers

If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.

AleksAgata [21]

Answer:

a) $f_1=5.587Hz$

b) $f_{n+1}-f_n=5.587Hz$

Explanation:

The frequency of the $n^{th}$ harmonic of a vibrating string of length L, linear density $\mu$ under a tension T is given by the formula:

$f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}$

a) So for the fundamental mode (n=1) we have, substituting our values:

$f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz$

b) The frequency difference between successive modes is the fundamental frequency, since:

$f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz$