All categories

3 years ago

A gas is confined in a cylinder with a piston. What happens when work is done on the gas.

Physics

2 answers:

goblinko [34]3 years ago

5 0

It combusts in to a vapor

yanalaym [24]3 years ago

3 0

The correct answer is:

The volume of the gas decreases

Explanation:

As the piston is forced down work is done on the gas. The force exercised by the gas pressure on the piston is in the adverse direction of the motion of the piston. The volume of the space the gas absorbs decreases. When we let go of the piston the gas extends. The force exercised by the gas is in the similar direction as the motion of the piston and work is done by the system.

You might be interested in

Same diagram... Which location shows SUMMER in the SOUTHERN hemisphere? *

LiRa [457]

Location 1 is correct

7 0

3 years ago

Read 2 more answers

A 35kg cannonball sits at rest on a flat level surface. What is the normal force exerted on the cannon ball. Use -9.8 m/s2 for a

Tom [10]

Answer:

a) N = 343 [N]

Explanation:

We must remember Newton's third law, which tells us that the force acting on a body is equal to the normal reaction force of equal magnitude but acting in the opposite direction.

$N=m*g$

where:

m = mass = 35 [kg]

g = gravity acceleration = 9.81 [m/s²]

$N = 35*9.8\\N= 343 [N]$

8 0

2 years ago

a spring is stretched 20 cm by a 30.0 n force. determine the work done in stretching the spring from 0 to 40 cm?

Alekssandra [29.7K]

The work done when a spring is stretched from 0 to 40cm is 4J.

What is work done?

Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.

The work done on the spring to stretch to 40cm is,

F = kx

where F is force, k is force constant.

k = F / x = 10 N / 20 * 10^-2 m = 50 N/m

W = 0.5 * k * (x)^2

where W = work done, k = force constant.

W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.

Therefore, the work done on the spring when it is stretched to 40cm is 4J.

To learn more about work done click on the given link brainly.com/question/25573309

#SPJ4

3 0

1 year ago

A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.

IrinaVladis [17]

Answer:

Approximately $722\; \rm m\cdot s^{-1}$ .

Explanation:

The average speed of a vehicle is calculated as:

$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$ .

In this question, the total distance is $2 \times 560\; \rm km = 1120\; \rm km$ .

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

$560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m$ .

$1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m$ .

Time required for the first part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s$ .

Time required for the second part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s$ .

The time required for the entire trip would be approximately $930 + 620 = 1550\; \rm s$ .

Calculate the average speed of this plane:

$\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}$ .

6 0

2 years ago

When the compound Na2S forms what happens to the Na and S ions

Mila [183]

Nothing, ATOMS lose or gain electrons to become IONS
<span>Two Na ATOMS lose 1 electron each and 1 S ATOM gains 2 electrons </span>

8 0

3 years ago

Read 2 more answers