All categories

2 years ago

600 J of work is done on a system in a process that decreases the thermal energy of the system by 300 J.

Physics

1 answer:

Daniel [21]2 years ago

6 0

Answer:

900 J

Explanation:

$\Delta U$ = Change in entropy = -300 J (Decrease)

$W$ = Work done = 600 J

$Q$ = Heat transferred

Change in internal energy is given by

$\Delta U=W+Q\\\Rightarrow Q=\Delta U-W\\\Rightarrow Q=-300-600\\\Rightarrow Q=-900\ J$

The heat transferred to or from the system as heat is -900 J.

900 J is transferred. As heat is transferred from the system the sign is negative.

You might be interested in

A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh

Novosadov [1.4K]

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

$\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}$ .....(I)

Where, $\delta t'$ = proper time

$\delta t$ = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

$\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}$

For the second interval,

Put the value of $\Delta t'$ in the equation (I)

$\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value in the equation

$\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}$

$\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}$

$\Delta t_{2}=-5.77\times10^{8}\ s$

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

8 0

3 years ago

An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it

steposvetlana [31]

Answer:

μs = 0.36

Explanation:

While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.
This acceleration is produced by the centripetal force.
Now, this force is not a different type of force, is the net force acting on the riders in this direction.
Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.
In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:

$F_{frmax} = \mu_{s}* F_{n} = m * g (1)$

where μs= coefficient of static friction (our unknown)

As we have already said Fn = Fc.
The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:

$F_{n} = m* \omega^{2} * r (2)$

Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:

$\mu_{s} = \frac{g}{\omega^{2} r} (3)$

Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:

$\omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)$

Replacing g, ω and r in (3):
$\mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)$

3 0

2 years ago

Light rays from stars bend toward smaller angles as they enter Earth's atmosphere. a. Explain why this happens using Snell's law

Grace [21]

Answer:

Following are the answer to this question:

Explanation:

In option (a):

The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.

In option (b):

Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.