Ask question

Ask question

All categories

Pachacha [2.7K]

4 years ago

12

A gas expands from an initial volume of 30.0 L to a final volume of 65.0 L at a constant pressure of 110kPa. How much work is do

ne by the gas?

1 answer:

Brilliant_brown [7]4 years ago

8 0

Answer:

3850 J

Explanation:

$V_{i}$ = initial volume of the gas = 30 L = 0.03 m³

$V_{f}$ = final volume of the gas = 65 L = 0.065 m³

$P$ = Constant pressure of the gas = 110 kPa = 110000 Pa

$W$ = Work done by the gas

Since the pressure is constant, Work done by the gas is given as

$W = P (V_{f} - V_{i})\\W = (110000) (0.065 - 0.03)\\W = 3850 J$

You might be interested in

The risks and expense of manned deep-water exploration must be balanced with____________.

ehidna [41]

Answer:

The risks and expense of manned deep water exploration must be balanced with the knowledge that will be gained.

Explanation:

6 0

3 years ago

Suppose you calculated the speed of light in an unknown substance to be 4.00x10^8 m/s. How could you tell if you made an error i

Mice21 [21]

Answer: You could tell if you made an error in your calculations by repeating the steps.

Speed of light is the fastest/maximum in vacuum which is equal to 3 × 10^8 m/s, therefore speed of light through any material equal to 4 × 10^8 m/s is physically and theoretically impossible and therefore incorrect.

Explanation:

6 0

2 years ago

A cup of hot coffee has been left to cool in a room with an ambient temperature of 21^\circ{\text{C}}21 ∘ C21, degrees, start te

Kay [80]

Answer:

Explanation:

From the equation given ; T(m)=21+74⋅10 −0.03m

Plugging the value of m = 10minutes for the time to calculate the temperature at that time instant. The detailed steps is as shown in the attached file

6 0

3 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

(c) 2.86s

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

<u>The answer for (a) is 1.6m</u>

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

<u>the answer to (b) is 6.86m</u>

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

and since $g=9.81m/s^2$

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

At $t=1.43s$ the car hits the water, so the horizontal distance at that time is:

$x(t)=x_{0}+v_{0}t$

$x(1.43)=(2m/s)(1.43s)=2.86m$

<u>the answer to (c) is 2.86s</u>

6 0

3 years ago

A volcano erupts, and a chunk of hot magma is launched horizontally with a speed of 205 m/s from a height of 550 m. We can ignor

Over [174]

Answer:

2170

Explanation:

the calcuations add up to it

7 0

3 years ago

Read 2 more answers