How much force would be required for a ten strand pulley system to lift 1000 newtons?

Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion

stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 240 K

= initial volume of gas =

= final volume of gas =

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

6 0

2 years ago

A car accelerates from rest at -3.00m/s^2. What is the velocity at the end of 5.0s? What is the displacement after5.0s?

Andrew [12]

Speed = (acceleration) x (time)
Velocity = (speed) in (direction of the speed)

Speed = (-3 m/s²) x (5 s) = 15 m/s
Velocity =
   (15 m/s) in the direction opposite to the direction you call positive.

Displacement = (distance between start-point and end-point)
   in the direction from start-point to end-point.

Distance = (1/2) (acceleration) (time)²
Distance = (1/2) (3 m/s²) (5 s)²
   = (1/2) (3 m/s²) (25 s²) = 37.5 meters

Displacement =
   37.5 meters in the direction opposite to the direction you call positive.

5 0

3 years ago

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0

3 years ago

A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficien

Liula [17]

Answer:

Yes it will move and a= 4.19m/s^2

Explanation:

In order for the box to move it needs to overcome the maximum static friction force

Max Static Friction = μFn(normal force)

plug in givens

Max Static friction = 31.9226

Since 36.6>31.9226, the box will move

Mass= Wieght/g which is 45.8/9.8= 4.67kg

Fnet = Fapp-Fk

= 36.6-16.9918

=19.6082

=ma

Solve for a=4.19m/s^2

7 0

3 years ago

Superman stops a "speeding locomotive" of mass 8.0x103 kg in 4.0 seconds. If the train was originally moving at 40.m/s,

madreJ [45]

Hello!

a) What is the change in momentum?

The change of the momentum is the velocity, because the velocity is reduced to zero.

We can calculate the aceleration, applicating the formula:

$\boxed{a=\frac{V-Vi}{t} }$

Like you see, the final velocity will be zero, so:

$a = \dfrac{0m/s-40m/s}{4s}$

$a = \dfrac{-40m/s}{4s}$

$a = -10\ m/s^{2}$

Like you see, the aceleration applicate by superman is of -10 m/s^2.

If the aceleration is negattive, means the velocity will decrease.

b.) What is the magnitud of the force wich superman exerts on the train?

For calculate the force applicate for superman, lets applicate the second law of Newton:

$\boxed{F=ma}$

Lets replace and resolve it:

F = 8x10^3 kg * (-10 m/s^2)

F = 8000 kg * (-10 m/s^2)

F = -80 000 N

The force applicated is of -80000 Newtons.

When the force is negative means the force applicated in the another direction of the object what is traveling.

8 0

3 years ago