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scoundrel [369]
3 years ago
12

Because the pressure falls, water boils at a lower temperature with increasing altitude. Consequently, cake mixes and boiled egg

s, among other foods, must be cooked different lengths of time. Determine the boiling temperature of water at 1000 and 2000 m elevation on a standard day, and compare with the sea-level value
Physics
1 answer:
AlexFokin [52]3 years ago
7 0

Answer:

1) The boiling point of water reduces by 3.28°C at 1,000 m above sea-level

2) The boiling point of water reduces by 6.56°C at 2,000 m above sea-level

Explanation:

The variation of the boiling point of water with elevation is given as follows

The boiling point reduces by 0.5°C for every 152.4 meter increase in elevation

At sea-level, the boiling point temperature of water = 100°C

1) At 1,000 m elevation, the boiling point temperature, T = 100 - (1,000/152.4) × 0.5 ≈ 96.72 °C

Therefore, the boiling point of water reduces by 100° - 96.72° = 3.28°C at 1,000 m above sea-level

2) At 2,000 m elevation, the boiling point temperature, T = 100 - (2,000/152.4) × 0.5 ≈ 93.44°C

The boiling point of water reduces by 100° - 93.44° = 6.56°C at 2,000 m above sea-level

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Hey! please help i’ll give brainliest!
Komok [63]

Question:

If you push a bowling ball and a golf ball with an equal force what will happen

Answer:

B)

Explanation:

Larger than the force used to push the object that has less mass. A golf ball and a bowling ball are moving at the same velocity. When gravity and air resistance are equal, the object has drawn its terminal velocity.

3 0
2 years ago
1. ¿Qué presión se ejerce sobre cada una de las cuatro patas de una mesa si su masa es de 20 kg y
mestny [16]
Tenemos.

Masa de la mesa = 20kg
Masa encima = 10kg
Masa total = 30kg
Area de cada pata = 20cm² = 20cm² * 1/10000cm² * 1m² =0,002m²

Presió(P)n que se ejerce sobre cada pata.

P =F/A
P = Masa por gravedad/A Masa = 30kg Gravedad =9,8m/s²
P =(30kg * 9,8m/s²)/0,002m²
P = 294kg * m/s²/0,002m² Pero kg *m/s² = Nw
P = 294Nw0,002m²
P = 147000 Nw/m² Pero Nw/m² = pa
p =1,47 * 10⁵ pa

Respuesta.
La presión que se ejerce sobre cada pata es de 1,47 * 10⁵pa
8 0
3 years ago
13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i
Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

8 0
2 years ago
A 7950-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s . A 2950-kg load, initially
Alchen [17]

The new speed of car is 10.9 m/s

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According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

m_1u_1 + m_2u_2 = (m_1 + m_2) v

v =\frac{m_1u_1}{m_1+m_2}

v = \frac{7950*15}{7950+2950}

v= 10.9 m/s

Therefore, the new speed of care is 10.9 m/s

Learn more about momentum here:

brainly.com/question/22257327

#SPJ1

5 0
2 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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