Ask question

Ask question

All categories

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

Guys plz help my C SHE IS ANNOYING I HAVE A TEST

solmaris [256]

Answer:

Compose a paragraph that identifies and describes each of the three Siberian regions.

Explanation:

8 0

3 years ago

En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

$Q=mc(T_2-T_1)$

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

$Q_1=-Q_2$

for water we have to

c = 4.18J / g ° C

replacing we have

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

$Q=mc(T_2-T_1)$

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

$Q_1=-Q_2$

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

7 0

3 years ago

Which country or region is coal-rich and oil-poor?

zimovet [89]

THE ANSWER FOR UR QUESTION IS

WHICH COUNTRY OR REGION IS COAL RICH AND OIL POOR
ANS:- UNITED STATES IS THE COUNTRY WHICH IS COAL RICH AND OIL POOR

HOPE THIS HELPS!!!

^_^ HAPPY TO HELP U ^_^

5 0

3 years ago

A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at

docker41 [41]

Answer:

$y = 4.36$

Explanation:

Let the height of the ladder be L

$L = 10$

Also:

Let $x = distance\ from\ the\ base\ of\ the\ ladder$

Let $y = height\ of\ the\ base\ of\ the\ ladder$

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

$L^2 = x^2 + y^2$ --- Pythagoras Theorem

When the base is 9ft from the wall, this means that:

$x = 9$

Substitute 9 for x and 10 for L in $L^2 = x^2 + y^2$

$10^2 = 9^2 + y^2$

$100 = 81 + y^2$

Make $y^2$ the subject

$y^2 = 100 - 81$

$y^2 = 19$

Make y the subject

$y = \sqrt{19$

$y = 4.36$

<em>Hence, the true distance at that point is approximately 4.36ft</em>

8 0

3 years ago

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0

3 years ago