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Montano1993 [528]

3 years ago

10

Which conditions are usually the effect of a low air pressure system? clear, dry weather cloudy, wet weather cold, dry weather h

ot, dry weather

1 answer:

sergejj [24]3 years ago

7 0

Your answer is hot and dry weather

You might be interested in

Importance of liquid pressure

trapecia [35]

Answer:

hope it helps...

Explanation:

Both the water in the ocean and the air in the atmosphere exert pressure because of their moving particles. ... This causes greater pressure. Denser fluids such as water exert more pressure than less dense fluids such as air. The particles of denser fluids are closer together, so there are more collisions in a given area.

7 0

3 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

8 0

3 years ago

a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q

Tatiana [17]

Answer:

a) the correct answer is 1 , b) x=0 F=0, a=0

x= 0.050 F= -7.5 N, a= -15 m/s²

x= 0.150 F= 22.5 N, a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values are from the equilibrium position of the spring.

Let's calculate the force

x = 0

F = 0

x = 0.050

F = - 150 0.050

F = - 7.50 N

x = 0.150

F = - 150 0.150

F = - 22.5 N

let's use Newton's second law to find the acceleration

F = m a

a = F / m

x = 0 m

a = 0

x = 0.050 m

a = -7.50 / 0.50

a = - 15 m / s²

x = 0.150 m

a = - 22.5 / 0.50

a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte la fuerza es dada por la fuerza de Hoke,

F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es 1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza

x=0

F= 0

x=0.050

F = - 150 0.050

F= - 7.50 N

x= 0.150

F= - 150 0.150

F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

F = m a

a = F/m

x =0 m

a = 0

x= 0.050 m

a = -7.50/ 0.50

a =- 15 m/s²

x= 0.150 m

a= - 22.5 / 0.50

a= - 45 m/s²

7 0

3 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

3 years ago

An engine converts 95% of its energy to mechanical. What happens to the other 5% of its energy?

Elza [17]

Answer:

Here, 5% of energy wasted as a heat lost or work done against frictional forces.

Explanation:

No engine has yet made that can convert its all input energy into mechanical energy. Some of the input energy goes in vain in the form of heat and friction.

6 0

3 years ago