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2 years ago

If the humidity in a room of volume 450 m3 at 30 ∘C is 75%, what mass of water can still evaporate from an open pan?

1 answer:

5 0

From tables,

SVP at 30°C = 4.24 kPa

From ideal gas expressions;
n = PV/RT = (4.24*1000*450)/(8.314*303) = 757.4 moles

Now, 75% of 757.4 moles will evaporate leaving 20%. Then, 25% of 757.5 moles...
25% of 757.4 moles = 25/100*757.4 = 189.35 moles
Mass of 189.35 moles = 189.35 moles*18 g/mol = 3408.3 g ≈ 3.4 kg

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At time t=0 , a cart is at x=10 m and has a velocity of 3 m/s in the −x -direction. The cart has a constant acceleration in the

DENIUS [597]

Answer:

Explanation:

The minimum magnitude of acceleration = 3 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 3 x 1²

= - 3 + 1.5

= - 1.5 m

position at t = 1 s

= 10 - 1.5

= 8.5 m

The maximum magnitude of acceleration = 6 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 6 x 1²

= - 3 + 3

= 0

position at t = 1 s

= 10 +0

= 10 m

So range of position is 8.5 m to 10 m .

7 0

1 year ago

A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its

Katyanochek1 [597]

Answer:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

$d_{x}(t)=dcos(\theta (t))$ (1)

d is the length of the minute hand (d=D/2)
D is the diameter of the clock
t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

$\theta (t)=\omega *t$ (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

$\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}$

Now, let's put this value on (2)

$\theta (t)=\frac{\pi}{30}*t$

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

I hope it helps you!

6 0

1 year ago

Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th

DENIUS [597]

Answer:

a) vboat = 5.95 m/s b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w = $\frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s$