Ask question

Ask question

All categories

2 years ago

7

A mixture of gas expands from 0.03m³ to 0.06m³ at constant pressure of 1MPa & absorbs energy of 84 J. Calculate the change i

n internal energy?

1 answer:

leonid [27]2 years ago

4 0

Answer:54 kj

Explanation:P1 = P2 = 1000 kPa

1Q2 = 84 kJ

1W2 = P1 (V2 – V1)

= 1000 (0.06 – 0.03) = 30 kJ

1Q2 = 1W2 + 1U2

U2 – U1= 1Q2 – 1W2 = 84 – 30 = 54 kJ

You might be interested in

I just need help with these seven!

Anna [14]

21. light changes its direction when travelling through a new medium because denser mediums have a higher angle of refraction.

5 0

3 years ago

Two postal delivery workers have different routes. They both travel from the post office, to neighborhoods to deliver mail, and

SOVA2 [1]

they travel the same distance

7 0

2 years ago

how would the velocity of the book change if the applied force were equal to the sliding friction force

Inessa [10]

Yes, Sliding friction opposes the movement of the book, slowing it down.sliding That's the 'kinetic' kind.. According to Newton's second law, F=ma. That is, the bear's acceleration should be proportional to the total force acting on the bear. As the bear's velocity is constant, its acceleration is zero. Therefore, the total Force acting on the bear is zero. Thus, the friction has to be equal in magnitude and opposite in direction to the bear's weight. As W=mg, we get that its weight is <span>9.8*400=3,920 Newton. Thus, the friction acting on the bear is 3,920 Newton</span>

3 0

3 years ago

The same wave travels for 2 seconds, what is its frequency?*

sveticcg [70]

Answer:

In waves distance is measured by wave length and time is measured by frequency or period.

velocity ratio=wave length multiply by frequency.

HENCE, if the same wave travels for 2 econds its frequency will be 2Hz.

Explanation:

8 0

2 years ago

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f

zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

$v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s$

$v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s$

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

$s = v_vt + gt^2/2$

$10 = 19.97t - 9.8t^2/2$

$4.9t^2 - 19.97t + 10 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}$

$t= \frac{19.9658877511823\pm14.24}{9.8}$

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

$s_1 = v_ht_1 = 15*0.58 = 8.8 m$

$s_2 = v_ht_2 = 15*3.49 = 52.5m$

8 0

2 years ago