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2 years ago

8

A certain amount of a monatomic gas is maintained at constant volume as it is cooled by 50K. This feat is accomplished by removi

ng 400 J of energy from the gas. How much work is done by the gas?
a) 400 J
b) -400 J
c) zero
d) none of the above

1 answer:

gtnhenbr [62]2 years ago

5 0

Answer:

The correct answer is option a. 400J

Explanation:

Hello!

Let's solve this!

To know if the work is positive or negative we have to know the following:

The work is positive when the gas performs (expansion)

The work is negative when done outside on the gas (compression)

In this case the work is done by the gas, so it is positive.

The correct answer is option a. 400J

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calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it

Lera25 [3.4K]

The force of F=10 N produces an extension of
$x=4 cm=0.04 m$
on the string, so the spring constant is equal to
$k= \frac{F}{x}= \frac{10 N}{0.04 m}=250 N/m$

Then the string is stretched by $\Delta x=40 cm=0.40 m$ . The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
$W= \Delta U= \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}(250 N/m)(0.40 m)^2=20 J$

5 0

2 years ago

A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of

vitfil [10]

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

3 0

2 years ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

2 years ago

11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0

Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0

2 years ago

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The <u><em>direction of the ball</em></u> is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

2 years ago