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The surface pressures at the bases of warm and cold columns of air are equal. air pressure in the warm column of air will ______

EastWind [94]

In atmospheric science, surface pressure is the atmospheric pressure at a location on Earth's surface. It is directly proportional to the mass of air over that location. For numerical reasons, atmospheric models such as general circulation models (GCMs) usually predict the nondimensional logarithm of surface pressure.

The answer is decrease more slowly

3 0

2 years ago

Fish, poultry, lean meats, and nuts should be consumed for which of the following nutrients? A. calcium B. carbohydrates C. prot

eimsori [14]

Answer:

The correct answers is option C, protein

Explanation:

Various food items (derived from plants and animals) are protein rich but some have protein higher than the others and thus these food items are considered as high protein rich food. For example - lean meat and skinless poultry, eggs , legumes, nuts and seeds etc. are all good source of proteins. These sources of food also have vitamin, calcium and carbohydrate but since they have more of protein than other food item thus they are considered good source of protein only. While several other food items are good sources of vitamin, calcium and carbohydrates.

8 0

2 years ago

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An object is located 13.5 cm in front of a convex mirror, the image being 7.05 cm behind the mirror. A second object, twice as t

goldfiish [28.3K]

Answer:

Second object is located at 42.03 cm in front of mirror

Explanation:

In this question we have given,

object distance from convex mirror ,u=-13.5cm

Image distance from convex mirror,v=7.05cm

let focal length of convex mirror be f

we have to find the distance of second object from convex mirror

we know that u, v and f are related by following formula

$\frac{1}{f} =\frac{1}{v}+ \frac{1}{u}$ .............(1)

put values of u and v in equation (1)

we got,

$\frac{1}{f} =\frac{1}{7.05}+ \frac{1}{-13.5}$

$\frac{1}{f}=\frac{13.5-7.05}{13.5\times 7.05}$

$\frac{1}{f}=\frac{6.45}{13.5\times 7.05}\\f=13.5\times 1.09\\f=14.75$

we have given that

second object is twice as tall as the first object

and image height of both objects are same

it means

$o_{2}=2o_{1}\\i_{1}=i_{2}$ .............(2)

we know that

$\frac{v}{u}=\frac{i}{o}\\i=\frac{o\times v}{u}$

therefore,

$i_{1}=\frac{o_{1}\times v}{u}$ .................(3)

put values of v and u in equation 3

$i_{1}=-\frac{o_{1}\times 7.05}{13.5}$

$i_{1}=-0.52o_{1}$

therefore from equation 2

$i_{2}=-0.52o_{1}$

we know that

$i_{2}=\frac{o_{2}\times V}{U}$ .................(4)

put value of $i_{2}$ and $o_{2}$ in equation 4

$-.52o_{1}=\frac{2o_{1}\times V}{U}$

$U=\frac{2o_{1}\times V}{-.52o_{1}} \\U=-3.85V$

we know that U,V and f are related by following formula

$\frac{1}{f} =\frac{1}{V}+ \frac{1}{U}$ .............(5)

put values of f and U in equation 5

we got

$\frac{1}{14.75} =\frac{1}{V}- \frac{1}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}\\V=\frac{2.85\times 14.75}{3.85}\\V=10.91 cm$

Therefore,

U=-10.91\times 3.85

U=-42.03 cm

Second object is located at 42.03 cm in front of mirror

4 0

2 years ago

An object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic friction is 0.20. If the plane i

hodyreva [135]

Answer: 9.312 m/s

Explanation:

The friction force (opposite to the motion) is Fa = μ*m*g*cos(α) with μ = kinetic friction. The force that makes the motion is

F = m*g*sin(α).

The Newton's law gives:

F - Fa = m*a

m*g*sin(α) - μ*m*g*cos(α) = m*a

g*sin(α) - μ*g*cos(α) = a so a = 4.335 m/s²

It's a uniformly accelerated motion:

Space

S = 0.5*a*t²

10 = 0.5*a*t²

=> t = 2.148 s

Velocity

V = a*t = 9.312 m/s.

5 0

2 years ago

A square block of steel with volume 10 cm3 and mass of 75 g is cut precisely in half. The density of the two smaller pieces is n

dem82 [27]

A. Density only depends on the substance. It doesn't matter whether you have a little chip of it or a supertanker full of it ... the density doesn't change.

4 0

2 years ago

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