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natima [27]
3 years ago
11

Line d represents water. if the atmospheric pressure in a flask is lowered to 70 kpa, water would boil at what temperature?

Physics
1 answer:
sp2606 [1]3 years ago
5 0
The diagram is in the picture attached.
Options are:
A) 32 °C
B) 70 °C
C) 92 °C
D) 100 °C

In order to find the value required, you need to look at the diagram and follow these steps:
1) search for the value of 70 kPa on the y-axis;
2) move on a horizontal line towards the right until you reach the line D;
3) move on a vertical line down, towards the x-axis;
4) read at what value of °C you are at.

Doing so, you can see that you are at a value a little bit above 90 °C (see picture).

Hence, the correct answer is C) 92°C.

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A. Angular momentum is always conserved would be the correct answer.

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7 0
3 years ago
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A body travel according to the law x(t)=2t+3t3 calculate the accelaration of body at t=2s
faust18 [17]

Answer:

28m/s

Explanation:

Given law,

⇒ x(t) = 2t + 3t³

Where, t = 2s

Then,

⇒ x(2) = 2(2) + 3(2)³

⇒ 4 + 3(8)

⇒ 4 + 24

⇒ 28 m/s

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3 years ago
A man's higher initial acceleration means that a man can outrun a horse over a very short race. A simple - but plausible - model
Lera25 [3.4K]

Answer:

 t_man = 10.16 s,   t_horse = 10.73 s,  the winner is the man

Explanation:

To solve this problem we are going to find the time of each one separately.

Man we look for distance and time during acceleration

         x₁ =  v₀ t₁ + ½ a₁ t₁²

as it comes out of rest its initial velocity is zero

        x₁ = ½ a₁ t₁²

        x₁ = ½ 6.0 1.8²

        x₁ = 9.72 m

at this point its speed is

        v₁ = v₀ + a t

        v₁ = 0 + 6  1.8

        v₁ = 10.8 m / s

From here on it continues at constant speed, the distance that the man needs to travel from the point where the man leaves at 100m is

        x₂ = 100 - x₁

        x₂ = 100- 9.72

        x₂ = 90.28 m

the time for this part is

        v₁ = x₂ / t₂

         t₂ = x₂ / v₂

         t₂ = 90.28 / 10.8

         t₂ = 8.36 s

the total time for the man is

        t_man = t₁ + t₂

        t _man = 1.8 + 8.36

        t_man = 10.16 s

We repeat the calculation for the horse

distance traveled during the acceleration period

         x₃ = v₀ t + ½ a₂ t₃²

as part of rest its initial velocity is zero

        x₃ = ½  a₂ t₃²

        x₃ = ½  5.0  4.8²

        x₃ = 57.6 m

the velocity at this point is

         v₃ = v₀ + a₂ t₃

         v₃ = 0 + 5  4.8

         v₃ = 24 m / s

the rest of the route is at constant speed, the remaining distance

         x₄ = 200 - x₃

         x₄ = 200 - 57.6

         x₄ = 142.4 m

the time to go through it is

         t₄ = x₄ / v₃

         t₄ = 142.4 / 24

         t₄ = 5.93 s

the total time for the horse is

         t_horse = t₃ + t₄

         t_horse = 4.8 + 5.93

         t_horse = 10.73 s

when we compare the times we see that the man arrives a little before the horse, the winner is the man

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