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2 years ago

At high pressures, what two factors will cause deviations during ideal gas law calculations?

1 answer:

4 0

At high pressures, the two factors that cause deviation during ideal gas law calculation are the size of molecular and intermolecular force.

The high pressure causes the molecules to approach each other at a very close distance. In that case, if the intermolecular force of attraction is high, the molecules may undergo a state transition, which will result in a completely different outcome as predicted by Ideal gas law.

If the size of the molecule is more, that is for heavy gases like refrigerants, the ideal gas law deviates due to the fact that, with increase in pressure, the volume of gas can no longer be considered as negligible.

To know more about intermolecular force go here

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APAKAH VOLUME 5 KG ES SAMA DENGAN VOLUME 5KG AIR APA YG MEMBEDAKAN KEDUA ZAT TERSEBUT

horrorfan [7]

Answer:c

Explanation:ganglandd

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4 years ago

A light-rail commuter train accelerates at a rate of 1.35 m/s2. How long does it take to reach its top speed of 80.0 km/h, start

Len [333]

Answer:

a) 16.46 seconds.

b) 13.46 seconds

c) 2.67 m/s²

Explanation:

Acceleration = a = 1.35 m/s²

Final velocity = v = 80 km/h = $80\frac{1000}{3600}=22.22\ m/s$

Initial velocity = u = 0

Equation of motion

$v=u+at\\\Rightarrow 22.22=0+1.35t\\\Rightarrow t=\frac{22.22}{1.35}=16.46\ s$

Time taken to accelerate to top speed is 16.46 seconds.

Acceleration = a = -1.65 m/s²

Initial velocity = u = 80 km/h= $80\frac{1000}{3600}=22.22\ m/s$

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=22.22-1.65t\\\Rightarrow t=\frac{22.22}{1.65}=13.46\ s$

Time taken to stop the train from top speed is 13.46 seconds

Initial velocity = u = 80 km/h= $80\frac{1000}{3600}=22.22\ m/s$

Time taken = t = 8.3 s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=22.22+a8.3\\\Rightarrow a=\frac{-22.22}{8.3}=-2.67\ m/s^2$

Emergency deceleration is 2.67 m/s²

8 0

4 years ago

Effciency of a lever is never 100% or more. why?Give reason

Troyanec [42]

Answer:

Ideally, the work output of a lever should match the work input. However, because of resistance, the output power is nearly always be less than the input power. As a result, the efficiency would go below $100\%$ .

Explanation:

In an ideal lever, the size of the input and output are inversely proportional to the distances between these two forces and the fulcrum. Let $D_\text{in}$ and $D_\text{out}$ denote these two distances, and let $F_\text{in}$ and $F_\text{out}$ denote the input and the output forces. If the lever is indeed idea, then:

$F_\text{in} \cdot D_\text{in} = F_\text{out} \cdot D_\text{out}$ .

Rearrange to obtain:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}$

Class two levers are levers where the perpendicular distance between the fulcrum and the input is greater than that between the fulcrum and the output. For this ideal lever, that means $D_\text{in} > D_\text{out}$ , such that $F_\text{in} < F_\text{out}$ .

Despite $F_\text{in} < F_\text{out}$ , the amount of work required will stay the same. Let $s_\text{out}$ denote the required linear displacement for the output force. At a distance of $D_\text{out}$ from the fulcrum, the angular displacement of the output force would be $\displaystyle \frac{s_\text{out}}{D_\text{out}}$ . Let $s_\text{in}$ denote the corresponding linear displacement required for the input force. Similarly, the angular displacement of the input force would be $\displaystyle \frac{s_\text{in}}{D_\text{in}}$ . Because both the input and the output are on the same lever, their angular displacement should be the same:

$\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}$ .

Rearrange to obtain:

$\displaystyle s_\text{in}=s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}$ .

While increasing $D_\text{in}$ reduce the size of the input force $F_\text{in}$ , doing so would also increase the linear distance of the input force $s_\text{in}$ . In other words, $F_\text{in}$ will have to move across a longer linear distance in order to move $F_\text{out}$ by the same $s_\text{out}$ .

The amount of work required depends on both the size of the force and the distance traveled. Let $W_\text{in}$ and $W_\text{out}$ denote the input and output work. For this ideal lever:

$\begin{aligned}W_\text{in} &= F_\text{in} \cdot s_\text{in} \\ &= \left(F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}\right) \cdot \left(s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}\right) \\ &= F_\text{out} \cdot s_\text{out} = W_\text{out}\end{aligned}$ .

In other words, the work input of the ideal lever is equal to the work output.

The efficiency of a machine can be measured as the percentage of work input that is converted to useful output. For this ideal lever, that ratio would be $100\%$ - not anything higher than that.

On the other hand, non-ideal levers take in more work than they give out. The reason is that because of resistance, $F_\text{in}$ would be larger than ideal:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}} + F(\text{resistance})$ .

As a result, in real (i.e., non-ideal) levers, the work input will exceed the useful work output. The efficiency will go below $100\%$ ,

4 0

3 years ago

C) m1: D) m2: E) r. Please help me I need help now

Nesterboy [21]

A) F = gravitational force

b) G = universal gravitational constant (6.67 × 10-11 N-m2/kg2)

c) m1 = mass of the body 1

d) m2 = mass of body 2

e) r = radius or distance between the two bodies.

Hope this helps!

6 0

3 years ago

The isotope 23893Np has a half-life of 2.0 days. If 4.00 grams are produced at noon on Monday, what will be the mass of neptuniu

viva [34]

Answer:

0.25 gram of neptunium is remaining

Explanation:

First we calculate the no. of half lives passed. For that we have formula:

n = t/T

where,

n = no. of half lives passed = ?

t = total time passed = 8 days (From Monday noon to Tuesday noon of following week)

T = Half Life Period = 2 days

Therefore,

n = 8 days/2 days

n = 4

Now, for the remaining mass of neptunium, we use the formula:

m = (mi)/(2)^n

where,

mi = initial mass of neptunium = 4.00 grams

m = remaining mass of neptunium = ?

Therefore,

m = 4 grams/2⁴

m = 0.25 gram

3 0

3 years ago