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3 years ago

What happens to air pressure as altitude decreases

2 answers:

8 0

Pressure decreases with increasing altitude. The pressure at any level in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels.

Mamont248 [21]3 years ago

5 0

Pressure with Height: pressure decreases with increasing altitude. The pressure at any level in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels.

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Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C.

zaharov [31]

Answer:

c. is more than that of the fluid.

Explanation:

This problem is based on the conservation of energy and the concept of thermal equilibrium

$heat= m s \Delta T
$

m= mass

s= specific heat

\DeltaT=change in temperature

let s1= specific heat of solid and s2= specific heat of liquid

then

Heat lost by solid= $20(s_1)(70-30)=800s_1
$

Heat gained by fluid= $100(s_2)(30-20)=1000s_2
$

Now heat gained = heat lost

therefore,

1000 S_2=800 S_1

S_1=1.25 S_2

so the specific heat of solid is more than that of the fluid.

8 0

3 years ago

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

3 years ago

What is the change in thermal energy E if the coefficent of kinetic friction between the box and floor is .4 , the distance the

Jet001 [13]

This question can be solved using the concept of friction energy.

The thermal energy change is b "258.4 J".

The change in thermal energy will be equal to the friction energy produced during the motion of the box.

$Change\ In\ Thermal\ Energy = E = Friction\ Energy\\\\E = \mu fd$

where,

μ = coefficient of kinetic friction = 0.4

f = force applied = 38 N

d = distance traveled by the box = 17 m

Therefore,

$E = (0.4)(38\ N)(17\ m)$

Learn more about friction energy here:

brainly.com/question/1343045?referrer=searchResults

7 0

2 years ago

A 4-kg cat is resting on a bookshelf that is 3m high. What is the cats gravitational potential energy relative to the floor if t

Kobotan [32]

Potential energy = mass x gravity x height

P.E = 4 x 9.8 x 3
P.E = 117.6 J

8 0

3 years ago

If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo

lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

temperature at t = 0

$T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}$

T(0) = 95°C

temperature after half hour of cooling

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

t = 30 minutes

$T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}$

$T( 30 ) = 20 + 75 \times 0.5488$

T(30) = 61.16° C

average of first half hour will be equal to

$T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt$

$T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30$

$T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30$

$T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]$

$T = \dfrac{1}{30}[600 - 2058.04 + 3750]$

T = 76.39°C

4 0

3 years ago